Class 10 Mathematics Chapter 8: Introduction to Trigonometry – Master NCERT Guide
1. SEO STRATEGY INTRODUCTION & CHAPTER MASTER OVERVIEW
Chapter 8, Introduction to Trigonometry, is one of the most vital units in Class 10 Mathematics under the CBSE curriculum. Trigonometry, derived from the Greek words trigonon (triangle) and metron (measure), is the branch of mathematics that explores relationships between the side lengths and angles of triangles. In Class 10, this study is restricted to right-angled triangles, laying the foundational groundwork for higher-level calculus, physics, engineering, and Class 11 Advanced Mathematics.
In the CBSE Class 10 Board Examination, Chapter 8 carries substantial weightage, typically accounting for 8 to 10 marks within the overall 12-mark Trigonometry unit (which includes Chapter 9: Some Applications of Trigonometry). Examiners frequently assess conceptual clarity through short numerical items, application-based multi-step proofs, and Higher Order Thinking Skills (HOTS) questions.
Core Conceptual Pillars
- Trigonometric Ratios: Defining the ratios of sides of a right-angled triangle with respect to an acute angle ($\sin \theta$, $\cos \theta$, $\tan \theta$, $\csc \theta$, $\sec \theta$, $\cot \theta$).
- Specific Angle Ratios: Evaluation of trigonometric ratios for standard angles: $0^\circ, 30^\circ, 45^\circ, 60^\circ,$ and $90^\circ$.
- Trigonometric Identities: Fundamental identities derived from the Pythagorean theorem, used to transform and simplify complex trigonometric expressions.
Common Student Misconceptions & Marking Scheme Pitfalls
- Misidentifying Opposite and Adjacent Sides: Students frequently confuse the “Opposite Side” (Perpendicular) and “Adjacent Side” (Base). The side opposite to the angle under consideration is always the perpendicular ($P$), while the side adjacent to it (other than the hypotenuse) is the base ($B$).
- Incorrect Trigonometric Notation: Writing $\sin \theta$ as $\sin \times \theta$ is a common conceptual error. $\sin \theta$ represents the sine function applied to angle $\theta$, not a multiplication of two terms.
- Identity Misapplication: Confusing $1 + \tan^2 \theta = \sec^2 \theta$ with $1 – \tan^2 \theta = \sec^2 \theta$ or squandering marks by writing $\sin^2 \theta + \cos^2 \theta = 2$ instead of $1$.
Master Summary Table: Ratios, Standard Angles & Identities
| Ratio / Angle | Formula Definition | 0∘ | 30∘ | 45∘ | 60∘ | 90∘ | Reciprocal Relation |
| $\sin \theta$ | $\frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{P}{H}$ | $0$ | $\frac{1}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{\sqrt{3}}{2}$ | $1$ | $\sin \theta = \frac{1}{\csc \theta}$ |
| $\cos \theta$ | $\frac{\text{Base}}{\text{Hypotenuse}} = \frac{B}{H}$ | $1$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{\sqrt{2}}$ | $\frac{1}{2}$ | $0$ | $\cos \theta = \frac{1}{\sec \theta}$ |
| $\tan \theta$ | $\frac{\text{Perpendicular}}{\text{Base}} = \frac{P}{B}$ | $0$ | $\frac{1}{\sqrt{3}}$ | $1$ | $\sqrt{3}$ | Not Defined | $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{1}{\cot \theta}$ |
| $\csc \theta$ | $\frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{H}{P}$ | Not Defined | $2$ | $\sqrt{2}$ | $\frac{2}{\sqrt{3}}$ | $1$ | $\csc \theta = \frac{1}{\sin \theta}$ |
| $\sec \theta$ | $\frac{\text{Hypotenuse}}{\text{Base}} = \frac{H}{B}$ | $1$ | $\frac{2}{\sqrt{3}}$ | $\sqrt{2}$ | $2$ | Not Defined | $\sec \theta = \frac{1}{\cos \theta}$ |
| $\cot \theta$ | $\frac{\text{Base}}{\text{Perpendicular}} = \frac{B}{P}$ | Not Defined | $\sqrt{3}$ | $1$ | $\frac{1}{\sqrt{3}}$ | $0$ | $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{1}{\tan \theta}$ |
Fundamental Trigonometric Identities:
- $\sin^2 \theta + \cos^2 \theta = 1$
- $1 + \tan^2 \theta = \sec^2 \theta \quad (\text{or } \sec^2 \theta – \tan^2 \theta = 1)$
- $1 + \cot^2 \theta = \csc^2 \theta \quad (\text{or } \csc^2 \theta – \cot^2 \theta = 1)$
2. IN-TEXT & SECTIONAL EXERCISES
Section 8.1: Trigonometric Ratios (Page No. 181)
Question 1. In $\Delta ABC$, right-angled at $B$, $AB = 24\text{ cm}$, $BC = 7\text{ cm}$. Determine:
(i) $\sin A$, $\cos A$
(ii) $\sin C$, $\cos C$
Answer:
Let us draw a right-angled triangle $\Delta ABC$ right-angled at $B$.
Given:
- Side adjacent to right angle $AB = 24\text{ cm}$
- Side adjacent to right angle $BC = 7\text{ cm}$
By applying the Pythagoras theorem in $\Delta ABC$:
$$\begin{aligned} AC^2 &= AB^2 + BC^2 \\ AC^2 &= (24)^2 + (7)^2 \\ AC^2 &= 576 + 49 \\ AC^2 &= 625 \\ AC &= \sqrt{625} = 25\text{ cm} \end{aligned}$$
(i) For acute angle $A$:
- $\text{Side opposite to angle } A = BC = 7\text{ cm}$ (Perpendicular)
- $\text{Side adjacent to angle } A = AB = 24\text{ cm}$ (Base)
- $\text{Hypotenuse } AC = 25\text{ cm}$
Using the definitions of sine and cosine ratios:
$$\begin{aligned} \sin A &= \frac{\text{Side opposite to angle } A}{\text{Hypotenuse}} = \frac{BC}{AC} = \underline{\frac{7}{25}} \\ \cos A &= \frac{\text{Side adjacent to angle } A}{\text{Hypotenuse}} = \frac{AB}{AC} = \underline{\frac{24}{25}} \end{aligned}$$
(ii) For acute angle $C$:
- $\text{Side opposite to angle } C = AB = 24\text{ cm}$ (Perpendicular)
- $\text{Side adjacent to angle } C = BC = 7\text{ cm}$ (Base)
- $\text{Hypotenuse } AC = 25\text{ cm}$
Using the definitions:
$$\begin{aligned} \sin C &= \frac{\text{Side opposite to angle } C}{\text{Hypotenuse}} = \frac{AB}{AC} = \underline{\frac{24}{25}} \\ \cos C &= \frac{\text{Side adjacent to angle } C}{\text{Hypotenuse}} = \frac{BC}{AC} = \underline{\frac{7}{25}} \end{aligned}$$
Question 2. In the given figure, find $\tan P – \cot R$.
Plaintext
+------------------------------------+
| Right-Angled Triangle PQR |
| |
| P |
| |\ |
| | \ |
| 12 \ 13 |
| | \ |
| |____\ |
| Q R |
| (90°) |
+------------------------------------+
Answer:
Given:
- In $\Delta PQR$, $\angle Q = 90^\circ$
- $PQ = 12\text{ cm}$
- $PR = 13\text{ cm}$
Applying the Pythagoras theorem in right-angled $\Delta PQR$:
$$\begin{aligned} PR^2 &= PQ^2 + QR^2 \\ (13)^2 &= (12)^2 + QR^2 \\ 169 &= 144 + QR^2 \\ QR^2 &= 169 – 144 = 25 \\ QR &= \sqrt{25} = 5\text{ cm} \end{aligned}$$
Now, calculating $\tan P$ (with respect to angle $P$):
$$\tan P = \frac{\text{Side opposite to angle } P}{\text{Side adjacent to angle } P} = \frac{QR}{PQ} = \frac{5}{12}$$
Calculating $\cot R$ (with respect to angle $R$):
$$\cot R = \frac{\text{Side adjacent to angle } R}{\text{Side opposite to angle } R} = \frac{QR}{PQ} = \frac{5}{12}$$
Now, substituting values into the required expression:
$$\begin{aligned} \tan P – \cot R &= \frac{5}{12} – \frac{5}{12} \\ &= \underline{0} \end{aligned}$$
Question 3. If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$. [BOARD EXAM FAVORITE]
Answer:
Given:
$$\sin A = \frac{3}{4}$$
By definition of the sine ratio in a right-angled triangle:
$$\sin A = \frac{\text{Side opposite to angle } A}{\text{Hypotenuse}}$$
Let the side opposite to angle $A$ be $3k$ and the hypotenuse be $4k$, where $k$ is a positive real constant (scaling factor).
By the Pythagoras theorem:
$$\begin{aligned} (\text{Hypotenuse})^2 &= (\text{Opposite})^2 + (\text{Adjacent})^2 \\ (4k)^2 &= (3k)^2 + (\text{Adjacent})^2 \\ 16k^2 &= 9k^2 + (\text{Adjacent})^2 \\ (\text{Adjacent})^2 &= 16k^2 – 9k^2 = 7k^2 \\ \text{Adjacent side} &= \sqrt{7k^2} = k\sqrt{7} \end{aligned}$$
Now, computing $\cos A$ and $\tan A$:
$$\begin{aligned} \cos A &= \frac{\text{Side adjacent to angle } A}{\text{Hypotenuse}} = \frac{k\sqrt{7}}{4k} = \underline{\frac{\sqrt{7}}{4}} \\ \tan A &= \frac{\text{Side opposite to angle } A}{\text{Side adjacent to angle } A} = \frac{3k}{k\sqrt{7}} = \underline{\frac{3}{\sqrt{7}}} \end{aligned}$$
Question 4. Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.
Answer:
The given relation can be rewritten as:
$$\cot A = \frac{8}{15}$$
By definition:
$$\cot A = \frac{\text{Side adjacent to angle } A}{\text{Side opposite to angle } A}$$
Let $\text{Adjacent side} = 8k$ and $\text{Opposite side} = 15k$, where $k > 0$.
Applying the Pythagoras theorem:
$$\begin{aligned} (\text{Hypotenuse})^2 &= (\text{Opposite})^2 + (\text{Adjacent})^2 \\ &= (15k)^2 + (8k)^2 \\ &= 225k^2 + 64k^2 \\ &= 289k^2 \\ \text{Hypotenuse} &= \sqrt{289k^2} = 17k \end{aligned}$$
Calculating $\sin A$ and $\sec A$:
$$\begin{aligned} \sin A &= \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{15k}{17k} = \underline{\frac{15}{17}} \\ \sec A &= \frac{\text{Hypotenuse}}{\text{Adjacent side}} = \frac{17k}{8k} = \underline{\frac{17}{8}} \end{aligned}$$
Question 5. Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.
Answer:
Given:
$$\sec \theta = \frac{13}{12} = \frac{\text{Hypotenuse}}{\text{Base}}$$
Let $\text{Hypotenuse } H = 13k$ and $\text{Base } B = 12k$, where $k > 0$.
Applying the Pythagoras theorem to find the Perpendicular ($P$):
$$\begin{aligned} H^2 &= P^2 + B^2 \\ (13k)^2 &= P^2 + (12k)^2 \\ 169k^2 &= P^2 + 144k^2 \\ P^2 &= 169k^2 – 144k^2 = 25k^2 \\ P &= \sqrt{25k^2} = 5k \end{aligned}$$
Now, determining all remaining five trigonometric ratios:
$$\begin{aligned} \sin \theta &= \frac{P}{H} = \frac{5k}{13k} = \underline{\frac{5}{13}} \\ \cos \theta &= \frac{B}{H} = \frac{12k}{13k} = \underline{\frac{12}{13}} \\ \tan \theta &= \frac{P}{B} = \frac{5k}{12k} = \underline{\frac{5}{12}} \\ \csc \theta &= \frac{H}{P} = \frac{13k}{5k} = \underline{\frac{13}{5}} \\ \cot \theta &= \frac{B}{P} = \frac{12k}{5k} = \underline{\frac{12}{5}} \end{aligned}$$
Question 6. If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$. [CBSE 2023]
Answer:
Consider two right-angled triangles $\Delta APC$ and $\Delta BQD$ such that $\angle C = 90^\circ$ and $\angle D = 90^\circ$, or consider a single triangle $\Delta ABC$ where $CD \perp AB$. Let us prove this using a single triangle $\Delta ABC$ with $CD \perp AB$ for simplicity and logical rigor.
Plaintext
+------------------------------------+
| Triangle ABC with CD perpendicular |
| to AB |
| |
| C |
| /|\ |
| / | \ |
| / | \ |
| / | \ |
| /____|____\ |
| A D B |
+------------------------------------+
In $\Delta ABC$, draw perpendicular $CD \perp AB$. Thus, $\Delta ADC$ and $\Delta BDC$ are right-angled triangles at $D$.
From $\Delta ADC$:
$$\cos A = \frac{AD}{AC}$$
From $\Delta BDC$:
$$\cos B = \frac{BD}{BC}$$
Given that $\cos A = \cos B$:
$$\frac{AD}{AC} = \frac{BD}{BC} = k \quad (\text{say, where } k > 0)$$
This gives:
$$AD = k \cdot AC \quad \text{and} \quad BD = k \cdot BC \quad \text{— (Equation 1)}$$
Applying Pythagoras theorem in $\Delta ADC$ and $\Delta BDC$:
In $\Delta ADC$: $CD^2 = AC^2 – AD^2$
In $\Delta BDC$: $CD^2 = BC^2 – BD^2$
Equating the two expressions for $CD^2$:
$$AC^2 – AD^2 = BC^2 – BD^2$$
Substituting values from Equation 1 into the left side:
$$\begin{aligned} AC^2 – (k \cdot AC)^2 &= BC^2 – (k \cdot BC)^2 \\ AC^2 (1 – k^2) &= BC^2 (1 – k^2) \end{aligned}$$
Dividing both sides by $(1 – k^2)$ (assuming $k \neq 1$; if $k=1$, $AC=AD$, which is impossible in a right triangle as hypotenuse $AC > AD$):
$$\begin{aligned} AC^2 &= BC^2 \\ AC &= BC \end{aligned}$$
Since sides opposite to equal angles in $\Delta ABC$ are equal ($AC = BC$):
$$\underline{\angle A = \angle B}$$
Hence proved.
Question 7. If $\cot \theta = \frac{7}{8}$, evaluate:
(i) $\frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)}$
(ii) $\cot^2 \theta$
Answer:
Given: $\cot \theta = \frac{7}{8}$.
(i) Consider the algebraic expansion of numerator and denominator using identity $(a+b)(a-b) = a^2 – b^2$:
$$E = \frac{(1 + \sin \theta)(1 – \sin \theta)}{(1 + \cos \theta)(1 – \cos \theta)} = \frac{1 – \sin^2 \theta}{1 – \cos^2 \theta}$$
Using fundamental trigonometric identities:
$$\sin^2 \theta + \cos^2 \theta = 1 \implies 1 – \sin^2 \theta = \cos^2 \theta \quad \text{and} \quad 1 – \cos^2 \theta = \sin^2 \theta$$
Substituting these into the expression:
$$E = \frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = (\cot \theta)^2$$
Given $\cot \theta = \frac{7}{8}$:
$$E = \left(\frac{7}{8}\right)^2 = \underline{\frac{49}{64}}$$
(ii) Evaluating $\cot^2 \theta$:
$$\cot^2 \theta = (\cot \theta)^2 = \left(\frac{7}{8}\right)^2 = \underline{\frac{49}{64}}$$
Question 8. If $3 \cot A = 4$, check whether $\frac{1 – \tan^2 A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A$ or not.
Answer:
Given:
$$3 \cot A = 4 \implies \cot A = \frac{4}{3}$$
Thus:
$$\tan A = \frac{1}{\cot A} = \frac{3}{4}$$
By definition, $\tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{4}$. Let $\text{Opposite} = 3k$ and $\text{Adjacent} = 4k$.
Applying the Pythagoras theorem:
$$\text{Hypotenuse} = \sqrt{(3k)^2 + (4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k$$
Hence:
$$\sin A = \frac{3k}{5k} = \frac{3}{5}, \quad \cos A = \frac{4k}{5k} = \frac{4}{5}$$
Evaluating Left-Hand Side (LHS):
$$\begin{aligned} \text{LHS} &= \frac{1 – \tan^2 A}{1 + \tan^2 A} = \frac{1 – \left(\frac{3}{4}\right)^2}{1 + \left(\frac{3}{4}\right)^2} \\ &= \frac{1 – \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{16 – 9}{16}}{\frac{16 + 9}{16}} = \frac{7/16}{25/16} = \frac{7}{25} \end{aligned}$$
Evaluating Right-Hand Side (RHS):
$$\begin{aligned} \text{RHS} &= \cos^2 A – \sin^2 A = \left(\frac{4}{5}\right)^2 – \left(\frac{3}{5}\right)^2 \\ &= \frac{16}{25} – \frac{9}{25} = \frac{7}{25} \end{aligned}$$
Since $\text{LHS} = \text{RHS} = \frac{7}{25}$, the equation is Yes, equal.
Question 9. In triangle $ABC$, right-angled at $B$, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of:
(i) $\sin A \cos C + \cos A \sin C$
(ii) $\cos A \cos C – \sin A \sin C$
Answer:
Given: $\Delta ABC$ right-angled at $B$, so $\angle B = 90^\circ$.
$$\tan A = \frac{1}{\sqrt{3}} = \frac{BC}{AB}$$
Let $BC = 1k$ and $AB = \sqrt{3}k$.
By the Pythagoras theorem:
$$\begin{aligned} AC^2 &= AB^2 + BC^2 \\ AC^2 &= (\sqrt{3}k)^2 + (1k)^2 \\ AC^2 &= 3k^2 + 1k^2 = 4k^2 \\ AC &= \sqrt{4k^2} = 2k \end{aligned}$$
Now calculate the ratios for angle $A$ and angle $C$:
- For angle $A$: $\sin A = \frac{BC}{AC} = \frac{1}{2}$, $\cos A = \frac{AB}{AC} = \frac{\sqrt{3}}{2}$
- For angle $C$: $\sin C = \frac{AB}{AC} = \frac{\sqrt{3}}{2}$, $\cos C = \frac{BC}{AC} = \frac{1}{2}$
(i) Evaluating $\sin A \cos C + \cos A \sin C$:
$$\begin{aligned} \sin A \cos C + \cos A \sin C &= \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \\ &= \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = \underline{1} \end{aligned}$$
(ii) Evaluating $\cos A \cos C – \sin A \sin C$:
$$\begin{aligned} \cos A \cos C – \sin A \sin C &= \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) – \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) \\ &= \frac{\sqrt{3}}{4} – \frac{\sqrt{3}}{4} = \underline{0} \end{aligned}$$
Question 10. In $\Delta PQR$, right-angled at $Q$, $PR + QR = 25\text{ cm}$ and $PQ = 5\text{ cm}$. Determine the values of $\sin P$, $\cos P$, and $\tan P$. [BOARD EXAM FAVORITE / CBSE 2023]
Answer:
Given:
- $\angle Q = 90^\circ$
- $PQ = 5\text{ cm}$
- $PR + QR = 25\text{ cm} \implies PR = 25 – QR$
Applying Pythagoras theorem in $\Delta PQR$:
$$\begin{aligned} PR^2 &= PQ^2 + QR^2 \\ (25 – QR)^2 &= 5^2 + QR^2 \\ (625 – 50QR + QR^2) &= 25 + QR^2 \end{aligned}$$
Subtracting $QR^2$ from both sides:
$$\begin{aligned} 625 – 50QR &= 25 \\ 50QR &= 625 – 25 \\ 50QR &= 600 \\ QR &= \frac{600}{50} = 12\text{ cm} \end{aligned}$$
Now, finding $PR$:
$$PR = 25 – QR = 25 – 12 = 13\text{ cm}$$
Evaluating trigonometric ratios for angle $P$:
- $\text{Perpendicular } (QR) = 12\text{ cm}$
- $\text{Base } (PQ) = 5\text{ cm}$
- $\text{Hypotenuse } (PR) = 13\text{ cm}$
$$\begin{aligned} \sin P &= \frac{QR}{PR} = \underline{\frac{12}{13}} \\ \cos P &= \frac{PQ}{PR} = \underline{\frac{5}{13}} \\ \tan P &= \frac{QR}{PQ} = \underline{\frac{12}{5}} \end{aligned}$$
Question 11. State whether the following statements are true or false. Justify your answer.
(i) The value of $\tan A$ is always less than 1.
Answer: False.
Justification: $\tan A = \frac{\text{Perpendicular}}{\text{Base}}$. In a right triangle, the perpendicular can be greater than the base. For instance, if $P = 12$ and $B = 5$, then $\tan A = \frac{12}{5} = 2.4 > 1$. Also, $\tan 60^\circ = \sqrt{3} \approx 1.732 > 1$.
(ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.
Answer: True.
Justification: $\sec A = \frac{\text{Hypotenuse}}{\text{Base}}$. Since hypotenuse is always the longest side in a right triangle, $\sec A$ must always be $\ge 1$. Here, $\frac{12}{5} = 2.4 \ge 1$, which is valid.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.
Answer: False.
Justification: $\cos A$ is the abbreviation for the cosine of angle $A$. The abbreviation used for cosecant is $\csc A$ (or $\text{cosec } A$).
(iv) $\cot A$ is the product of $\cot$ and $A$.
Answer: False.
Justification: $\cot A$ is a single composite term indicating the cotangent function applied to the angle $A$. Separated from $A$, ‘$\cot$’ has no independent mathematical meaning.
(v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.
Answer: False.
Justification: $\sin \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}}$. Since the hypotenuse is the longest side, $\text{Perpendicular} \le \text{Hypotenuse}$, so $\sin \theta$ can never exceed $1$. Here $\frac{4}{3} = 1.33… > 1$, which is impossible.
3. CHAPTER-END EXERCISES & NUMERICAL DRILLS
Section 8.2: Trigonometric Ratios of Specific Angles (Page No. 187)
Question 1. Evaluate the following:
(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
Answer:
Substituting exact standard values:
$\sin 60^\circ = \frac{\sqrt{3}}{2}$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 30^\circ = \frac{1}{2}$, $\cos 60^\circ = \frac{1}{2}$
$$\begin{aligned} \text{Value} &= \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) \\ &= \frac{3}{4} + \frac{1}{4} = \frac{4}{4} = \underline{1} \end{aligned}$$
(ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$
Answer:
Substituting values:
$\tan 45^\circ = 1$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$
$$\begin{aligned} \text{Value} &= 2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 – \left(\frac{\sqrt{3}}{2}\right)^2 \\ &= 2(1) + \frac{3}{4} – \frac{3}{4} = \underline{2} \end{aligned}$$
(iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \csc 30^\circ}$
Answer:
Substituting values:
$\cos 45^\circ = \frac{1}{\sqrt{2}}$, $\sec 30^\circ = \frac{2}{\sqrt{3}}$, $\csc 30^\circ = 2$
$$\begin{aligned} \text{Value} &= \frac{\frac{1}{\sqrt{2}}}{\frac{2}{\sqrt{3}} + 2} = \frac{\frac{1}{\sqrt{2}}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} \\ &= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}(\sqrt{3} + 1)} \end{aligned}$$
Rationalizing the denominator by multiplying numerator and denominator by $\sqrt{2}(\sqrt{3} – 1)$:
$$\begin{aligned} \text{Value} &= \frac{\sqrt{3} \cdot \sqrt{2}(\sqrt{3} – 1)}{2\sqrt{2} \cdot \sqrt{2} (\sqrt{3} + 1)(\sqrt{3} – 1)} \\ &= \frac{\sqrt{6}(\sqrt{3} – 1)}{4(3 – 1)} = \frac{\sqrt{18} – \sqrt{6}}{4(2)} \\ &= \underline{\frac{3\sqrt{2} – \sqrt{6}}{8}} \end{aligned}$$
(iv) $\frac{\sin 30^\circ + \tan 45^\circ – \csc 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
Answer:
Substituting values:
$\sin 30^\circ = \frac{1}{2}$, $\tan 45^\circ = 1$, $\csc 60^\circ = \frac{2}{\sqrt{3}}$, $\sec 30^\circ = \frac{2}{\sqrt{3}}$, $\cos 60^\circ = \frac{1}{2}$, $\cot 45^\circ = 1$
$$\text{Numerator} = \frac{1}{2} + 1 – \frac{2}{\sqrt{3}} = \frac{3}{2} – \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} – 4}{2\sqrt{3}}$$
$$\text{Denominator} = \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{2}{\sqrt{3}} + \frac{3}{2} = \frac{4 + 3\sqrt{3}}{2\sqrt{3}}$$
Dividing Numerator by Denominator:
$$\text{Value} = \frac{3\sqrt{3} – 4}{3\sqrt{3} + 4}$$
Rationalizing by multiplying top and bottom by $(3\sqrt{3} – 4)$:
$$\begin{aligned} \text{Value} &= \frac{(3\sqrt{3} – 4)^2}{(3\sqrt{3})^2 – (4)^2} = \frac{(3\sqrt{3})^2 + (4)^2 – 2(3\sqrt{3})(4)}{27 – 16} \\ &= \frac{27 + 16 – 24\sqrt{3}}{11} = \underline{\frac{43 – 24\sqrt{3}}{11}} \end{aligned}$$
(v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$
Answer:
Using $\sin^2 30^\circ + \cos^2 30^\circ = 1$ in the denominator:
$$\text{Denominator} = \left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$$
Evaluating the Numerator:
$$\begin{aligned} \text{Numerator} &= 5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 – (1)^2 \\ &= 5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) – 1 \\ &= \frac{5}{4} + \frac{16}{3} – 1 \\ &= \frac{15 + 64 – 12}{12} = \underline{\frac{67}{12}} \end{aligned}$$
Question 2. Choose the correct option and justify your choice:
(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} =$
(A) $\sin 60^\circ$
(B) $\cos 60^\circ$
(C) $\tan 60^\circ$
(D) $\sin 30^\circ$
Answer: (A) $\sin 60^\circ$
Justification:
$$\begin{aligned} \frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ} &= \frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}} \\ &= \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} = \sin 60^\circ \end{aligned}$$
(ii) $\frac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$
(A) $\tan 90^\circ$
(B) $1$
(C) $\sin 45^\circ$
(D) $0$
Answer: (D) $0$
Justification:
Since $\tan 45^\circ = 1$:
$$\frac{1 – (1)^2}{1 + (1)^2} = \frac{1 – 1}{1 + 1} = \frac{0}{2} = 0$$
(iii) $\sin 2A = 2 \sin A$ is true when $A =$
(A) $0^\circ$
(B) $30^\circ$
(C) $45^\circ$
(D) $60^\circ$
Answer: (A) $0^\circ$
Justification:
Substitute $A = 0^\circ$:
$\text{LHS} = \sin(2 \times 0^\circ) = \sin 0^\circ = 0$
$\text{RHS} = 2 \sin 0^\circ = 2(0) = 0$
$\text{LHS} = \text{RHS}$.
(iv) $\frac{2 \tan 30^\circ}{1 – \tan^2 30^\circ} =$
(A) $\cos 60^\circ$
(B) $\sin 60^\circ$
(C) $\tan 60^\circ$
(D) $\sin 30^\circ$
Answer: (C) $\tan 60^\circ$
Justification:
$$\begin{aligned} \frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 – \left(\frac{1}{\sqrt{3}}\right)^2} &= \frac{\frac{2}{\sqrt{3}}}{1 – \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \\ &= \frac{2}{\sqrt{3}} \times \frac{3}{2} = \sqrt{3} = \tan 60^\circ \end{aligned}$$
Question 3. If $\tan(A + B) = \sqrt{3}$ and $\tan(A – B) = \frac{1}{\sqrt{3}}$; $0^\circ < A + B \le 90^\circ$; $A > B$, find $A$ and $B$. [BOARD EXAM FAVORITE / CBSE 2022]
Answer:
Given:
- $\tan(A + B) = \sqrt{3}$Since $\tan 60^\circ = \sqrt{3}$, we get:
$$A + B = 60^\circ \quad \text{— (Equation 1)}$$
- $\tan(A – B) = \frac{1}{\sqrt{3}}$Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we get:
$$A – B = 30^\circ \quad \text{— (Equation 2)}$$
Adding Equation 1 and Equation 2:
$$\begin{aligned} (A + B) + (A – B) &= 60^\circ + 30^\circ \\ 2A &= 90^\circ \\ A &= \frac{90^\circ}{2} = \underline{45^\circ} \end{aligned}$$
Substituting $A = 45^\circ$ into Equation 1:
$$\begin{aligned} 45^\circ + B &= 60^\circ \\ B &= 60^\circ – 45^\circ = \underline{15^\circ} \end{aligned}$$
Verification: $A = 45^\circ > B = 15^\circ$, and $A + B = 60^\circ \le 90^\circ$. Satisfies all conditions.
Question 4. State whether the following are true or false. Justify your answer.
(i) $\sin(A + B) = \sin A + \sin B$.
Answer: False.
Justification: Let $A = 30^\circ$ and $B = 60^\circ$.
$\text{LHS} = \sin(30^\circ + 60^\circ) = \sin 90^\circ = 1$.
$\text{RHS} = \sin 30^\circ + \sin 60^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2}$.
$\text{LHS} \neq \text{RHS}$.
(ii) The value of $\sin \theta$ increases as $\theta$ increases.
Answer: True.
Justification: For $0^\circ \le \theta \le 90^\circ$, values are: $\sin 0^\circ = 0$, $\sin 30^\circ = 0.5$, $\sin 45^\circ \approx 0.707$, $\sin 60^\circ \approx 0.866$, $\sin 90^\circ = 1$. It increases continuously from $0$ to $1$.
(iii) The value of $\cos \theta$ increases as $\theta$ increases.
Answer: False.
Justification: For $0^\circ \le \theta \le 90^\circ$, values are: $\cos 0^\circ = 1$, $\cos 30^\circ \approx 0.866$, $\cos 45^\circ \approx 0.707$, $\cos 60^\circ = 0.5$, $\cos 90^\circ = 0$. It decreases from $1$ to $0$.
(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
Answer: False.
Justification: $\sin \theta = \cos \theta$ is true only for $\theta = 45^\circ$ ($\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$). For $\theta = 30^\circ$, $\sin 30^\circ = \frac{1}{2} \neq \cos 30^\circ = \frac{\sqrt{3}}{2}$.
(v) $\cot A$ is not defined for $A = 0^\circ$.
Answer: True.
Justification: $\cot 0^\circ = \frac{\cos 0^\circ}{\sin 0^\circ} = \frac{1}{0}$, which is division by zero, hence Not Defined.
Section 8.3: Trigonometric Identities (Page No. 193-194)
Question 1. Express the trigonometric ratios $\sin A$, $\sec A$, and $\tan A$ in terms of $\cot A$.
Answer:
- Expressing $\sin A$ in terms of $\cot A$:Using the identity: $1 + \cot^2 A = \csc^2 A$
$$\begin{aligned} \csc A &= \sqrt{1 + \cot^2 A} \\ \sin A &= \frac{1}{\csc A} = \underline{\frac{1}{\sqrt{1 + \cot^2 A}}} \end{aligned}$$
- Expressing $\tan A$ in terms of $\cot A$:By fundamental reciprocal identity:
$$\tan A = \underline{\frac{1}{\cot A}}$$
- Expressing $\sec A$ in terms of $\cot A$:Using the identity: $\sec^2 A = 1 + \tan^2 A$
$$\begin{aligned} \sec A &= \sqrt{1 + \tan^2 A} = \sqrt{1 + \left(\frac{1}{\cot A}\right)^2} \\ &= \sqrt{1 + \frac{1}{\cot^2 A}} = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}} = \underline{\frac{\sqrt{\cot^2 A + 1}}{\cot A}} \end{aligned}$$
Question 2. Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.
Answer:
- $\cos A$:
$$\cos A = \underline{\frac{1}{\sec A}}$$
- $\sin A$:Using $\sin^2 A + \cos^2 A = 1 \implies \sin A = \sqrt{1 – \cos^2 A}$:
$$\sin A = \sqrt{1 – \left(\frac{1}{\sec A}\right)^2} = \sqrt{1 – \frac{1}{\sec^2 A}} = \underline{\frac{\sqrt{\sec^2 A – 1}}{\sec A}}$$
- $\tan A$:Using $1 + \tan^2 A = \sec^2 A \implies \tan^2 A = \sec^2 A – 1$:
$$\tan A = \underline{\sqrt{\sec^2 A – 1}}$$
- $\csc A$:Reciprocal of $\sin A$:
$$\csc A = \frac{1}{\sin A} = \underline{\frac{\sec A}{\sqrt{\sec^2 A – 1}}}$$
- $\cot A$:Reciprocal of $\tan A$:
$$\cot A = \frac{1}{\tan A} = \underline{\frac{1}{\sqrt{\sec^2 A – 1}}}$$
Question 3. Choose the correct option. Justify your choice.
(i) $9 \sec^2 A – 9 \tan^2 A =$
(A) $1$
(B) $9$
(C) $8$
(D) $0$
Answer: (B) $9$
Justification:
Factoring out $9$:
$$9(\sec^2 A – \tan^2 A)$$
Using identity $\sec^2 A – \tan^2 A = 1$:
$$9(1) = 9$$
(ii) $(1 + \tan \theta + \sec \theta)(1 + \cot \theta – \csc \theta) =$
(A) $0$
(B) $1$
(C) $2$
(D) $-1$
Answer: (C) $2$
Justification:
Convert all ratios into $\sin \theta$ and $\cos \theta$:
$$\begin{aligned} \text{Term 1} &= 1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} = \frac{\cos \theta + \sin \theta + 1}{\cos \theta} \\ \text{Term 2} &= 1 + \frac{\cos \theta}{\sin \theta} – \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta – 1}{\sin \theta} \end{aligned}$$
Multiplying Term 1 and Term 2:
$$\begin{aligned} \text{Product} &= \frac{[(\sin \theta + \cos \theta) + 1][(\sin \theta + \cos \theta) – 1]}{\sin \theta \cos \theta} \\ &= \frac{(\sin \theta + \cos \theta)^2 – (1)^2}{\sin \theta \cos \theta} \\ &= \frac{(\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta) – 1}{\sin \theta \cos \theta} \end{aligned}$$
Substitute $\sin^2 \theta + \cos^2 \theta = 1$:
$$\text{Product} = \frac{1 + 2\sin \theta \cos \theta – 1}{\sin \theta \cos \theta} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = 2$$
(iii) $(\sec A + \tan A)(1 – \sin A) =$
(A) $\sec A$
(B) $\sin A$
(C) $\csc A$
(D) $\cos A$
Answer: (D) $\cos A$
Justification:
$$\begin{aligned} (\sec A + \tan A)(1 – \sin A) &= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)(1 – \sin A) \\ &= \left(\frac{1 + \sin A}{\cos A}\right)(1 – \sin A) \\ &= \frac{(1 + \sin A)(1 – \sin A)}{\cos A} = \frac{1 – \sin^2 A}{\cos A} \end{aligned}$$
Using $1 – \sin^2 A = \cos^2 A$:
$$\frac{\cos^2 A}{\cos A} = \cos A$$
(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$
(A) $\sec^2 A$
(B) $-1$
(C) $\cot^2 A$
(D) $\tan^2 A$
Answer: (D) $\tan^2 A$
Justification:
Using identities $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \csc^2 A$:
$$\frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$
Question 4. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) $(\csc \theta – \cot \theta)^2 = \frac{1 – \cos \theta}{1 + \cos \theta}$ [CBSE 2023]
Answer:
Evaluating Left-Hand Side (LHS):
$$\text{LHS} = (\csc \theta – \cot \theta)^2$$
Convert into $\sin \theta$ and $\cos \theta$:
$$\begin{aligned} \text{LHS} &= \left(\frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta}\right)^2 \\ &= \left(\frac{1 – \cos \theta}{\sin \theta}\right)^2 = \frac{(1 – \cos \theta)^2}{\sin^2 \theta} \end{aligned}$$
Substitute $\sin^2 \theta = 1 – \cos^2 \theta = (1 – \cos \theta)(1 + \cos \theta)$:
$$\begin{aligned} \text{LHS} &= \frac{(1 – \cos \theta)^2}{(1 – \cos \theta)(1 + \cos \theta)} \\ &= \frac{1 – \cos \theta}{1 + \cos \theta} = \text{RHS} \end{aligned}$$
Hence proved.
(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$ [BOARD EXAM FAVORITE / CBSE 2024]
Answer:
Evaluating Left-Hand Side (LHS):
$$\text{LHS} = \frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A}$$
Taking LCM of denominators:
$$\begin{aligned} \text{LHS} &= \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A} \\ &= \frac{\cos^2 A + (1 + \sin^2 A + 2\sin A)}{(1 + \sin A)\cos A} \end{aligned}$$
Regrouping $(\sin^2 A + \cos^2 A = 1)$:
$$\begin{aligned} \text{LHS} &= \frac{(\sin^2 A + \cos^2 A) + 1 + 2\sin A}{(1 + \sin A)\cos A} \\ &= \frac{1 + 1 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2 + 2\sin A}{(1 + \sin A)\cos A} \\ &= \frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} \\ &= 2 \sec A = \text{RHS} \end{aligned}$$
Hence proved.
(iii) $\frac{\tan \theta}{1 – \cot \theta} + \frac{\cot \theta}{1 – \tan \theta} = 1 + \sec \theta \csc \theta$ [CBSE 2023]
Answer:
Evaluating Left-Hand Side (LHS):
Convert $\tan \theta$ and $\cot \theta$ to $\sin \theta$ and $\cos \theta$:
$$\text{LHS} = \frac{\frac{\sin \theta}{\cos \theta}}{1 – \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 – \frac{\sin \theta}{\cos \theta}}$$
Simplify inner denominators:
$$\begin{aligned} \text{LHS} &= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta – \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta – \sin \theta}{\cos \theta}} \\ &= \frac{\sin^2 \theta}{\cos \theta(\sin \theta – \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta – \sin \theta)} \end{aligned}$$
Rewrite $(\cos \theta – \sin \theta) = -(\sin \theta – \cos \theta)$:
$$\text{LHS} = \frac{\sin^2 \theta}{\cos \theta(\sin \theta – \cos \theta)} – \frac{\cos^2 \theta}{\sin \theta(\sin \theta – \cos \theta)}$$
Combine fractions over common denominator $\sin \theta \cos \theta (\sin \theta – \cos \theta)$:
$$\text{LHS} = \frac{\sin^3 \theta – \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta – \cos \theta)}$$
Apply algebraic identity $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$:
$$\begin{aligned} \text{LHS} &= \frac{(\sin \theta – \cos \theta)(\sin^2 \theta + \sin \theta \cos \theta + \cos^2 \theta)}{\sin \theta \cos \theta (\sin \theta – \cos \theta)} \\ &= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} \quad (\text{since } \sin^2 \theta + \cos^2 \theta = 1) \\ &= \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta} \\ &= \csc \theta \sec \theta + 1 = 1 + \sec \theta \csc \theta = \text{RHS} \end{aligned}$$
Hence proved.
(iv) $\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$
Answer:
Evaluating Left-Hand Side (LHS):
$$\text{LHS} = \frac{1 + \sec A}{\sec A} = \frac{1}{\sec A} + \frac{\sec A}{\sec A} = \cos A + 1 = 1 + \cos A$$
Evaluating Right-Hand Side (RHS):
$$\text{RHS} = \frac{\sin^2 A}{1 – \cos A}$$
Use identity $\sin^2 A = 1 – \cos^2 A = (1 – \cos A)(1 + \cos A)$:
$$\text{RHS} = \frac{(1 – \cos A)(1 + \cos A)}{1 – \cos A} = 1 + \cos A$$
Since $\text{LHS} = \text{RHS} = 1 + \cos A$, the identity is proved.
(v) Prove that $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \csc A + \cot A$, using identity $\csc^2 A = 1 + \cot^2 A$. [BOARD EXAM FAVORITE / CBSE 2023]
Answer:
Evaluating Left-Hand Side (LHS):
$$\text{LHS} = \frac{\cos A – \sin A + 1}{\cos A + \sin A – 1}$$
Divide every term in numerator and denominator by $\sin A$:
$$\text{LHS} = \frac{\frac{\cos A}{\sin A} – \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} – \frac{1}{\sin A}} = \frac{\cot A – 1 + \csc A}{\cot A + 1 – \csc A} = \frac{(\cot A + \csc A) – 1}{(\cot A – \csc A) + 1}$$
Substitute $1 = \csc^2 A – \cot^2 A = (\csc A – \cot A)(\csc A + \cot A)$ in numerator:
$$\begin{aligned} \text{LHS} &= \frac{(\csc A + \cot A) – (\csc^2 A – \cot^2 A)}{\cot A – \csc A + 1} \\ &= \frac{(\csc A + \cot A) – (\csc A + \cot A)(\csc A – \cot A)}{\cot A – \csc A + 1} \end{aligned}$$
Factor out $(\csc A + \cot A)$ from the numerator:
$$\begin{aligned} \text{LHS} &= \frac{(\csc A + \cot A) [1 – (\csc A – \cot A)]}{\cot A – \csc A + 1} \\ &= \frac{(\csc A + \cot A) (1 – \csc A + \cot A)}{(1 – \csc A + \cot A)} \\ &= \csc A + \cot A = \text{RHS} \end{aligned}$$
Hence proved.
(vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$ [CBSE 2024]
Answer:
Evaluating Left-Hand Side (LHS):
$$\text{LHS} = \sqrt{\frac{1 + \sin A}{1 – \sin A}}$$
Multiply numerator and denominator inside the square root by $(1 + \sin A)$:
$$\begin{aligned} \text{LHS} &= \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 – \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 – \sin^2 A}} \\ &= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} \\ &= \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS} \end{aligned}$$
Hence proved.
(vii) $\frac{\sin \theta – 2 \sin^3 \theta}{2 \cos^3 \theta – \cos \theta} = \tan \theta$ [CBSE 2023]
Answer:
Evaluating Left-Hand Side (LHS):
Factor out $\sin \theta$ from numerator and $\cos \theta$ from denominator:
$$\text{LHS} = \frac{\sin \theta (1 – 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta – 1)}$$
Since $\sin^2 \theta + \cos^2 \theta = 1$, express $\cos^2 \theta$ in terms of $\sin^2 \theta$:
$$2 \cos^2 \theta – 1 = 2(1 – \sin^2 \theta) – 1 = 2 – 2\sin^2 \theta – 1 = 1 – 2\sin^2 \theta$$
Substituting this into denominator:
$$\begin{aligned} \text{LHS} &= \frac{\sin \theta (1 – 2\sin^2 \theta)}{\cos \theta (1 – 2\sin^2 \theta)} \\ &= \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS} \end{aligned}$$
Hence proved.
(viii) $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$ [BOARD EXAM FAVORITE / CBSE 2023, 2024]
Answer:
Evaluating Left-Hand Side (LHS):
Expand using $(a+b)^2 = a^2 + 2ab + b^2$:
$$\text{LHS} = (\sin^2 A + 2\sin A \csc A + \csc^2 A) + (\cos^2 A + 2\cos A \sec A + \sec^2 A)$$
Substitute $\sin A \csc A = 1$ and $\cos A \sec A = 1$:
$$\text{LHS} = \sin^2 A + 2(1) + \csc^2 A + \cos^2 A + 2(1) + \sec^2 A$$
Regroup $(\sin^2 A + \cos^2 A = 1)$:
$$\begin{aligned} \text{LHS} &= (\sin^2 A + \cos^2 A) + 2 + 2 + \csc^2 A + \sec^2 A \\ &= 1 + 4 + \csc^2 A + \sec^2 A = 5 + \csc^2 A + \sec^2 A \end{aligned}$$
Substitute standard identity transformations $\csc^2 A = 1 + \cot^2 A$ and $\sec^2 A = 1 + \tan^2 A$:
$$\begin{aligned} \text{LHS} &= 5 + (1 + \cot^2 A) + (1 + \tan^2 A) \\ &= 5 + 1 + 1 + \tan^2 A + \cot^2 A \\ &= 7 + \tan^2 A + \cot^2 A = \text{RHS} \end{aligned}$$
Hence proved.
(ix) $(\csc A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}$
Answer:
Evaluating Left-Hand Side (LHS):
$$\text{LHS} = \left(\frac{1}{\sin A} – \sin A\right)\left(\frac{1}{\cos A} – \cos A\right)$$
Simplify inside brackets:
$$\text{LHS} = \left(\frac{1 – \sin^2 A}{\sin A}\right)\left(\frac{1 – \cos^2 A}{\cos A}\right) = \left(\frac{\cos^2 A}{\sin A}\right)\left(\frac{\sin^2 A}{\cos A}\right) = \sin A \cos A$$
Evaluating Right-Hand Side (RHS):
$$\text{RHS} = \frac{1}{\tan A + \cot A} = \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{1}{\frac{1}{\sin A \cos A}} = \sin A \cos A$$
Since $\text{LHS} = \text{RHS} = \sin A \cos A$, the identity is proved.
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 – \tan A}{1 – \cot A}\right)^2 = \tan^2 A$
Answer:
Part 1: Evaluate $\frac{1 + \tan^2 A}{1 + \cot^2 A}$:
$$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\csc^2 A} = \frac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$
Part 2: Evaluate $\left(\frac{1 – \tan A}{1 – \cot A}\right)^2$:
$$\left(\frac{1 – \tan A}{1 – \frac{1}{\tan A}}\right)^2 = \left(\frac{1 – \tan A}{\frac{\tan A – 1}{\tan A}}\right)^2 = \left(\frac{-( \tan A – 1) \cdot \tan A}{\tan A – 1}\right)^2 = (-\tan A)^2 = \tan^2 A$$
Thus, Part 1 = Part 2 = $\tan^2 A$. Hence proved.
4. GRAMMAR, VOCABULARY & EXTENDED PRACTICAL APPLICATIONS
In scientific, surveying, and engineering domains, mathematical language requires absolute clarity. Below is the structural terminology framework used in geometric trigonometry.
Plaintext
+-------------------------------------------------------------------------+
| TRIGONOMETRIC TERMINOLOGY & FORMULAE |
+-------------------------------------------------------------------------+
| Term | Structural Mathematical Meaning |
+--------------------+----------------------------------------------------+
| Hypotenuse | Longest side of a right triangle, always opposite |
| | to the 90-degree angle. |
| Perpendicular | Side directly opposite to the acute reference |
| | angle theta. |
| Base | Side adjacent to the reference angle theta (other |
| | than the hypotenuse). |
| Angle of Elevation | Angle formed between horizontal line of sight and |
| | line of sight up to an object above observer. |
| Identity | An equality relation true for all admissible |
| | values of the contained variables. |
+-------------------------------------------------------------------------+
Real-World Practical Drill: Heights & Inaccessible Heights
Scenario: A civil engineer uses a total station device to measure the slope distance and vertical incline angle of a proposed bridge support post. The measured right-angled triangle has a base of $30\text{ m}$ along the ground, and the angle of elevation to the top of the post is $30^\circ$. Calculate the height of the post and the exact length of the structural support beam (hypotenuse).
Solution:
Let $AB$ be the height of the post (Perpendicular $P$), $BC = 30\text{ m}$ be the ground distance (Base $B$), and $AC$ be the beam (Hypotenuse $H$).
- Finding Height ($P$):
$$\begin{aligned} \tan 30^\circ &= \frac{AB}{BC} \\ \frac{1}{\sqrt{3}} &= \frac{AB}{30} \\ AB &= \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = \underline{10\sqrt{3}\text{ m} \approx 17.32\text{ m}} \end{aligned}$$
- Finding Beam Length ($H$):
$$\begin{aligned} \cos 30^\circ &= \frac{BC}{AC} \\ \frac{\sqrt{3}}{2} &= \frac{30}{AC} \\ AC &= \frac{60}{\sqrt{3}} = \underline{20\sqrt{3}\text{ m} \approx 34.64\text{ m}} \end{aligned}$$
5. 15 HIGH-YIELD FREQUENTLY ASKED QUESTIONS (BOARD EXAM FAQs)
Q1. If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$, show that $\cos \theta – \sin \theta = \sqrt{2} \sin \theta$. [CBSE HOTS]
Answer:
Given: $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$.
Squaring both sides:
$$\begin{aligned} (\sin \theta + \cos \theta)^2 &= (\sqrt{2} \cos \theta)^2 \\ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta &= 2\cos^2 \theta \end{aligned}$$
Rearranging terms:
$$\begin{aligned} 2\sin \theta \cos \theta &= 2\cos^2 \theta – \cos^2 \theta – \sin^2 \theta \\ 2\sin \theta \cos \theta &= \cos^2 \theta – \sin^2 \theta \end{aligned}$$
Now, consider the expression $(\cos \theta – \sin \theta)^2$:
$$(\cos \theta – \sin \theta)^2 = \cos^2 \theta + \sin^2 \theta – 2\sin \theta \cos \theta = 1 – 2\sin \theta \cos \theta$$
Substitute $2\sin \theta \cos \theta = 2\cos^2 \theta – 1$ (from $\sin^2 \theta + \cos^2 \theta = 1$ rearranged):
$$\begin{aligned} (\cos \theta – \sin \theta)^2 &= 1 – (2\cos^2 \theta – 1) = 2 – 2\cos^2 \theta = 2(1 – \cos^2 \theta) = 2\sin^2 \theta \\ \cos \theta – \sin \theta &= \sqrt{2\sin^2 \theta} = \underline{\sqrt{2} \sin \theta} \end{aligned}$$
Hence proved.
Q2. If $\tan \theta + \sin \theta = m$ and $\tan \theta – \sin \theta = n$, prove that $m^2 – n^2 = 4\sqrt{mn}$. [CBSE HOTS]
Answer:
Given: $m = \tan \theta + \sin \theta$, $n = \tan \theta – \sin \theta$.
Evaluating LHS ($m^2 – n^2$):
$$\begin{aligned} m^2 – n^2 &= (m + n)(m – n) \\ &= [(\tan \theta + \sin \theta) + (\tan \theta – \sin \theta)] [(\tan \theta + \sin \theta) – (\tan \theta – \sin \theta)] \\ &= (2\tan \theta)(2\sin \theta) = \underline{4\tan \theta \sin \theta} \end{aligned}$$
Evaluating RHS ($4\sqrt{mn}$):
$$\begin{aligned} mn &= (\tan \theta + \sin \theta)(\tan \theta – \sin \theta) = \tan^2 \theta – \sin^2 \theta \\ &= \frac{\sin^2 \theta}{\cos^2 \theta} – \sin^2 \theta = \sin^2 \theta \left(\frac{1}{\cos^2 \theta} – 1\right) \\ &= \sin^2 \theta (\sec^2 \theta – 1) = \sin^2 \theta \tan^2 \theta \end{aligned}$$
Taking the square root:
$$4\sqrt{mn} = 4\sqrt{\sin^2 \theta \tan^2 \theta} = \underline{4\sin \theta \tan \theta}$$
Since $\text{LHS} = \text{RHS} = 4\tan \theta \sin \theta$, the identity holds true.
Q3. If $\sec \theta + \tan \theta = p$, express $\sin \theta$ in terms of $p$. [CBSE 2023]
Answer:
Given:
$$\sec \theta + \tan \theta = p \quad \text{— (Equation 1)}$$
Using identity $\sec^2 \theta – \tan^2 \theta = 1 \implies (\sec \theta + \tan \theta)(\sec \theta – \tan \theta) = 1$:
$$\sec \theta – \tan \theta = \frac{1}{p} \quad \text{— (Equation 2)}$$
Adding Equation 1 and Equation 2:
$$2\sec \theta = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \sec \theta = \frac{p^2 + 1}{2p}$$
Subtracting Equation 2 from Equation 1:
$$2\tan \theta = p – \frac{1}{p} = \frac{p^2 – 1}{p} \implies \tan \theta = \frac{p^2 – 1}{2p}$$
Now calculate $\sin \theta = \frac{\tan \theta}{\sec \theta}$:
$$\sin \theta = \frac{\frac{p^2 – 1}{2p}}{\frac{p^2 + 1}{2p}} = \underline{\frac{p^2 – 1}{p^2 + 1}}$$
Q4. Assertion-Reason Question:
Assertion (A): For any acute angle $\theta$, $\sin^2 \theta + \cos^2 \theta = 1$.
Reason (R): $\sec^2 \theta – \tan^2 \theta = 1$ is valid for all $0^\circ \le \theta < 90^\circ$.
(A) Both A and R are true, and R is the correct explanation of A.
(B) Both A and R are true, but R is NOT the correct explanation of A.
(C) A is true, but R is false.
(D) A is false, but R is true.
Answer: (B) Both A and R are true, but R is NOT the correct explanation of A.
Explanation: Both statements are independently true foundational identities. However, R does not explain why $\sin^2 \theta + \cos^2 \theta = 1$ (which stems directly from the Pythagorean theorem).
Q5. If $x = a \cos \theta$ and $y = b \sin \theta$, find the value of $b^2 x^2 + a^2 y^2$.
Answer:
Substitute $x$ and $y$ into the given expression:
$$\begin{aligned} b^2 x^2 + a^2 y^2 &= b^2(a \cos \theta)^2 + a^2(b \sin \theta)^2 \\ &= b^2 a^2 \cos^2 \theta + a^2 b^2 \sin^2 \theta \\ &= a^2 b^2 (\cos^2 \theta + \sin^2 \theta) \end{aligned}$$
Since $\sin^2 \theta + \cos^2 \theta = 1$:
$$a^2 b^2 (1) = \underline{a^2 b^2}$$
Q6. Evaluate without standard table lookup: $\frac{\sin 30^\circ + \tan 45^\circ}{\sec 60^\circ}$.
Answer:
Substitute known standard values: $\sin 30^\circ = \frac{1}{2}$, $\tan 45^\circ = 1$, $\sec 60^\circ = 2$.
$$\text{Value} = \frac{\frac{1}{2} + 1}{2} = \frac{\frac{3}{2}}{2} = \underline{\frac{3}{4}}$$
Q7. If $\sin(A – B) = \frac{1}{2}$ and $\cos(A + B) = \frac{1}{2}$, find $A$ and $B$.
Answer:
Since $\sin 30^\circ = \frac{1}{2}$:
$$A – B = 30^\circ \quad \text{— (1)}$$
Since $\cos 60^\circ = \frac{1}{2}$:
$$A + B = 60^\circ \quad \text{— (2)}$$
Adding (1) and (2): $2A = 90^\circ \implies A = 45^\circ$.
Substituting in (2): $45^\circ + B = 60^\circ \implies B = 15^\circ$.
Q8. Case-Study Question: Trigonometric Ratios in Ramp Design
A civil contractor is constructing a wheelchair ramp for a hospital entrance. The ramp top must reach a height of $3\text{ m}$ ($AB$). The angle of inclination $\theta$ must satisfy $\sin \theta = 0.6$.
Plaintext
+------------------------------------+
| Ramp Structure Diagram |
| |
| A |
| |\ |
| | \ |
|3m| \ Ramp |
| | \ |
| |____\ |
| B C |
| (90°) |
+------------------------------------+
(i) Find the required length of the ramp ($AC$).
(ii) Find the horizontal base distance ($BC$).
Answer:
(i) Given $\sin \theta = \frac{AB}{AC} = 0.6 = \frac{3}{5}$.
$$\frac{3}{AC} = \frac{3}{5} \implies AC = \underline{5\text{ m}}$$
(ii) Applying Pythagoras theorem to find base $BC$:
$$BC = \sqrt{AC^2 – AB^2} = \sqrt{5^2 – 3^2} = \sqrt{25 – 9} = \sqrt{16} = \underline{4\text{ m}}$$
Q9. Prove that $\frac{1 + \cos \theta – \sin^2 \theta}{\sin \theta (1 + \cos \theta)} = \cot \theta$.
Answer:
Substitute $\sin^2 \theta = 1 – \cos^2 \theta$ into the numerator:
$$\begin{aligned} \text{LHS} &= \frac{1 + \cos \theta – (1 – \cos^2 \theta)}{\sin \theta (1 + \cos \theta)} \\ &= \frac{(1 + \cos \theta) – (1 – \cos \theta)(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} \\ &= \frac{(1 + \cos \theta)[1 – (1 – \cos \theta)]}{\sin \theta (1 + \cos \theta)} = \frac{1 – 1 + \cos \theta}{\sin \theta} = \frac{\cos \theta}{\sin \theta} = \underline{\cot \theta} \end{aligned}$$
Hence proved.
Q10. If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta – b \cos \theta = n$, prove that $a^2 + b^2 = m^2 + n^2$.
Answer:
Squaring $m$ and $n$ and adding:
$$\begin{aligned} m^2 + n^2 &= (a \cos \theta + b \sin \theta)^2 + (a \sin \theta – b \cos \theta)^2 \\ &= (a^2 \cos^2 \theta + b^2 \sin^2 \theta + 2ab \sin \theta \cos \theta) + (a^2 \sin^2 \theta + b^2 \cos^2 \theta – 2ab \sin \theta \cos \theta) \\ &= a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) \\ &= a^2(1) + b^2(1) = \underline{a^2 + b^2} \end{aligned}$$
Hence proved.
Q11. If $\csc \theta + \cot \theta = k$, prove that $\cos \theta = \frac{k^2 – 1}{k^2 + 1}$.
Answer:
Express in $\sin \theta$ and $\cos \theta$:
$$\frac{1 + \cos \theta}{\sin \theta} = k \implies \sin \theta = \frac{1 + \cos \theta}{k}$$
Squaring both sides and using $\sin^2 \theta = 1 – \cos^2 \theta$:
$$\begin{aligned} 1 – \cos^2 \theta &= \frac{(1 + \cos \theta)^2}{k^2} \\ (1 – \cos \theta)(1 + \cos \theta) &= \frac{(1 + \cos \theta)^2}{k^2} \end{aligned}$$
Dividing by $(1 + \cos \theta)$ assuming $\cos \theta \neq -1$:
$$k^2 (1 – \cos \theta) = 1 + \cos \theta \implies k^2 – k^2 \cos \theta = 1 + \cos \theta \implies \cos \theta(k^2 + 1) = k^2 – 1 \implies \underline{\cos \theta = \frac{k^2 – 1}{k^2 + 1}}$$
Hence proved.
Q12. If $\tan A = 1$ and $\tan B = \sqrt{3}$, evaluate $\cos A \cos B – \sin A \sin B$.
Answer:
Since $\tan A = 1 \implies A = 45^\circ$.
Since $\tan B = \sqrt{3} \implies B = 60^\circ$.
$$\begin{aligned} \text{Value} &= \cos 45^\circ \cos 60^\circ – \sin 45^\circ \sin 60^\circ \\ &= \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right) – \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) = \underline{\frac{1 – \sqrt{3}}{2\sqrt{2}}} \end{aligned}$$
Q13. Prove that $\frac{\tan A + \sec A – 1}{\tan A – \sec A + 1} = \frac{1 + \sin A}{\cos A}$.
Answer:
Use identity $1 = \sec^2 A – \tan^2 A$ in numerator:
$$\begin{aligned} \text{LHS} &= \frac{(\tan A + \sec A) – (\sec^2 A – \tan^2 A)}{\tan A – \sec A + 1} \\ &= \frac{(\sec A + \tan A)[1 – (\sec A – \tan A)]}{\tan A – \sec A + 1} = \frac{(\sec A + \tan A)(\tan A – \sec A + 1)}{\tan A – \sec A + 1} \\ &= \sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \underline{\frac{1 + \sin A}{\cos A}} \end{aligned}$$
Hence proved.
Q14. Is $\sin(60^\circ + 30^\circ) = \sin 60^\circ + \sin 30^\circ$?
Answer: No.
$\text{LHS} = \sin 90^\circ = 1$.
$\text{RHS} = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \approx 1.366 \neq 1$.
Q15. Simplify: $(1 + \tan^2 \theta)(1 – \sin \theta)(1 + \sin \theta)$.
Answer:
Apply identity $(1 – \sin \theta)(1 + \sin \theta) = 1 – \sin^2 \theta = \cos^2 \theta$:
$$\text{Expression} = (1 + \tan^2 \theta)(\cos^2 \theta)$$
Apply $1 + \tan^2 \theta = \sec^2 \theta$:
$$\sec^2 \theta \cdot \cos^2 \theta = \left(\frac{1}{\cos^2 \theta}\right) \cdot \cos^2 \theta = \underline{1}$$
6. CONCLUDING BOARD TOPPER STRATEGY
To score a perfect 100% in CBSE Board Examination Trigonometry questions, follow these examiner-backed structural guidelines:
- Explicit Triangle Conventions: Always sketch a clear, labeled right-angled triangle diagram in the margin for ratio problems. Label the reference angle $\theta$, Hypotenuse ($H$), Perpendicular ($P$), and Base ($B$).
- Mandatory Step Structure: Never skip stating the fundamental identity used in your transformations. Write
[Using identity: sin²θ + cos²θ = 1]in bold inline brackets alongside the step. - Rationalization & Formats: Always rationalize radical terms in final denominators (e.g., convert $\frac{3}{\sqrt{2}}$ to $\frac{3\sqrt{2}}{2}$).
- Identity Proof Protocol: Work strictly from LHS to RHS or solve both sides to a common mathematical expression independently. Clearly highlight the final line with $\text{LHS} = \text{RHS}$, followed by Hence Proved.