NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry — Free PDF

This chapter introduces the six trigonometric ratios (sin, cos, tan, cosec, sec, cot) for acute angles in a right triangle. You will learn their values at standard angles ($0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$), the complementary angle relationships, and the three fundamental trigonometric identities.

The chapter has 4 exercises covering:
- Exercise 8.1: Trigonometric ratios — finding all six ratios given one ratio or side lengths
- Exercise 8.2: Values at standard angles — evaluating expressions using the standard angle table
- Exercise 8.3: Complementary angles — using $\sin(90^\circ - \theta) = \cos\theta$ and related results
- Exercise 8.4: Trigonometric identities — proving identities using the three fundamental identities

This chapter typically carries 6-10 marks in the board exam, often combined with Chapter 9 (Some Applications of Trigonometry). Identity proofs are the most challenging part and frequently appear as 3-4 mark questions. Mastering this chapter requires memorising the standard angle table and practising identity manipulation until the patterns become second nature.

Trigonometric Ratios (for angle $\theta$ in a right triangle):

Also: $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$.

Standard Angle Values:

**Memory trick for $\sin$ values:** Write $\dfrac{\sqrt{0}}{2}, \dfrac{\sqrt{1}}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \dfrac{\sqrt{4}}{2}$ which gives $0, \dfrac{1}{2}, \dfrac{1}{\sqrt{2}}, \dfrac{\sqrt{3}}{2}, 1$. The $\cos$ values are the same in reverse order.

Complementary Angle Relationships:

Three Fundamental Trigonometric Identities:

Useful rearrangements:
- From (I): $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$
- From (II): $\sec^2\theta - \tan^2\theta = 1$, which factors as $(\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1$
- From (III): $\csc^2\theta - \cot^2\theta = 1$, which factors as $(\csc\theta - \cot\theta)(\csc\theta + \cot\theta) = 1$

**Problem 1: If $\tan A = \dfrac{4}{3}$, find the other trigonometric ratios of angle $A$.**

Solution:
$\tan A = \dfrac{4}{3}$, so opposite $= 4k$, adjacent $= 3k$.

By Pythagoras: hypotenuse $= \sqrt{(4k)^2 + (3k)^2} = \sqrt{16k^2 + 9k^2} = 5k$.

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**Problem 2: If $\sin A = \dfrac{3}{4}$, find $\cos A$ and $\tan A$.**

Solution:
$\sin A = \dfrac{3}{4}$: opposite $= 3k$, hypotenuse $= 4k$.

Adjacent $= \sqrt{(4k)^2 - (3k)^2} = \sqrt{16k^2 - 9k^2} = k\sqrt{7}$.

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**Problem 3: In $\triangle ABC$ right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Find $\sin A$, $\cos A$, $\sin C$, $\cos C$.**

Solution:

For angle $A$ (opposite $= BC = 7$, adjacent $= AB = 24$):

For angle $C$ (opposite $= AB = 24$, adjacent $= BC = 7$):

Notice: $\sin A = \cos C$ and $\cos A = \sin C$ (complementary angle relationship, since $A + C = 90^\circ$).

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**Problem 4: If $\sec\theta = \dfrac{13}{5}$, find all other trigonometric ratios.**

Solution:
$\sec\theta = \dfrac{13}{5}$, so $\cos\theta = \dfrac{5}{13}$. Adjacent $= 5k$, hypotenuse $= 13k$.

Opposite $= \sqrt{(13k)^2 - (5k)^2} = \sqrt{169k^2 - 25k^2} = 12k$.

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**Problem 5: If $\cot\theta = \dfrac{7}{8}$, evaluate: (i) $\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$, (ii) $\cot^2\theta$.**

Solution:
$\cot\theta = \dfrac{7}{8}$: adjacent $= 7k$, opposite $= 8k$, hypotenuse $= \sqrt{49k^2 + 64k^2} = k\sqrt{113}$.

(i) Using the identity $(1+a)(1-a) = 1 - a^2$:

(ii) $\cot^2\theta = \dfrac{49}{64}$.

**Problem 1: Evaluate $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$.**

Solution:

Answer: $2$.

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**Problem 2: Find $A$ if $\sin(A + 36^\circ) = \cos A$, where $(A + 36^\circ)$ is acute.**

Solution:
Using $\cos A = \sin(90^\circ - A)$:

Answer: $A = 27^\circ$.

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**Problem 3: Evaluate $\dfrac{\tan 65^\circ}{\cot 25^\circ}$.**

Solution:
$\cot 25^\circ = \tan(90^\circ - 25^\circ) = \tan 65^\circ$.

Answer: $1$.

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**Problem 4: Evaluate $\sin 18^\circ / \cos 72^\circ$.**

Solution:
$\cos 72^\circ = \sin(90^\circ - 72^\circ) = \sin 18^\circ$.

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**Problem 5: If $\tan 2A = \cot(A - 18^\circ)$ where $2A$ is an acute angle, find $A$.**

Solution:
$\cot(A - 18^\circ) = \tan(90^\circ - (A - 18^\circ)) = \tan(108^\circ - A)$.

Answer: $A = 36^\circ$.

**Problem 1: Evaluate $\dfrac{\cos 37^\circ}{\sin 53^\circ}$.**

Solution:
Since $37^\circ + 53^\circ = 90^\circ$, these are complementary angles.
$\sin 53^\circ = \cos(90^\circ - 53^\circ) = \cos 37^\circ$.

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**Problem 2: Show that $\tan 48^\circ \tan 23^\circ \tan 42^\circ \tan 67^\circ = 1$.**

Solution:
Pair up complementary angles:
$\tan 48^\circ \times \tan 42^\circ = \tan 48^\circ \times \cot 48^\circ = 1$ (since $42 + 48 = 90$)

$\tan 23^\circ \times \tan 67^\circ = \tan 23^\circ \times \cot 23^\circ = 1$ (since $23 + 67 = 90$)

Product $= 1 \times 1 = 1$.

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**Problem 3: If $\sec 4A = \csc(A - 20^\circ)$ where $4A$ is acute, find $A$.**

Solution:
$\sec 4A = \csc(90^\circ - 4A)$.

Answer: $A = 22^\circ$.

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**Problem 4: Without using tables, evaluate $\sin^2 20^\circ + \sin^2 70^\circ + \sin^2 35^\circ + \sin^2 55^\circ$.**

Solution:
$\sin^2 70^\circ = \cos^2 20^\circ$ (since $70 + 20 = 90$)
$\sin^2 55^\circ = \cos^2 35^\circ$ (since $55 + 35 = 90$)

Answer: $2$.

**Problem 1: Prove $(\csc\theta - \cot\theta)^2 = \dfrac{1 - \cos\theta}{1 + \cos\theta}$.**

Solution:
LHS:

Using $\sin^2\theta = 1 - \cos^2\theta = (1 - \cos\theta)(1 + \cos\theta)$:

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**Problem 2: Prove $(\sin A + \csc A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$.**

Solution:
LHS:

Note: $\sin A \cdot \csc A = 1$ and $\cos A \cdot \sec A = 1$.

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**Problem 3: Prove $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$.**

Solution:

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**Problem 4: Prove $\dfrac{\cos A}{1 + \sin A} + \dfrac{1 + \sin A}{\cos A} = 2\sec A$.**

Solution:
LHS: Take LCM:

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**Problem 5: Prove $\dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \sec\theta\csc\theta$.**

Solution:
Convert to $\sin$ and $\cos$:

Using $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$:

**Example 1: If $\sin\theta + \cos\theta = \sqrt{2}$, find $\tan\theta + \cot\theta$.**

Solution:
Square both sides: $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 2$.
$1 + 2\sin\theta\cos\theta = 2 \implies \sin\theta\cos\theta = \dfrac{1}{2}$.

Answer: $\tan\theta + \cot\theta = 2$.

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**Example 2: Prove that $\dfrac{1}{\sec A - \tan A} - \dfrac{1}{\cos A} = \dfrac{1}{\cos A} - \dfrac{1}{\sec A + \tan A}$.**

Solution:
Rearranging: $\dfrac{1}{\sec A - \tan A} + \dfrac{1}{\sec A + \tan A} = \dfrac{2}{\cos A}$.

LHS of this rearranged form:

This equals the RHS. Hence proved. $\blacksquare$

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**Example 3: Evaluate $\dfrac{5\cos^2 60^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$.**

Solution:
Numerator: $5\left(\dfrac{1}{2}\right)^2 + 4\left(\dfrac{2}{\sqrt{3}}\right)^2 - (1)^2 = 5 \times \dfrac{1}{4} + 4 \times \dfrac{4}{3} - 1 = \dfrac{5}{4} + \dfrac{16}{3} - 1$

$= \dfrac{15 + 64 - 12}{12} = \dfrac{67}{12}$

Denominator: $\sin^2 30^\circ + \cos^2 30^\circ = 1$ (by identity).

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**Example 4: If $7\sin^2\theta + 3\cos^2\theta = 4$, find $\tan\theta$.**

Solution:
$7\sin^2\theta + 3\cos^2\theta = 4$
$7\sin^2\theta + 3(1 - \sin^2\theta) = 4$
$7\sin^2\theta + 3 - 3\sin^2\theta = 4$
$4\sin^2\theta = 1$
$\sin^2\theta = \dfrac{1}{4} \implies \sin\theta = \dfrac{1}{2}$ (since $\theta$ is acute)

So $\theta = 30^\circ$ and $\tan 30^\circ = \dfrac{1}{\sqrt{3}}$.

Answer: $\tan\theta = \dfrac{1}{\sqrt{3}}$.

Mistake 1: Confusing which side is opposite and which is adjacent.
The opposite and adjacent sides depend on which angle you are considering. For angle $A$ in a right triangle, the side directly facing $A$ is the opposite, and the side next to $A$ (that is not the hypotenuse) is the adjacent. If you switch to angle $C$, the roles of opposite and adjacent swap.

**Mistake 2: Writing $\sin^2\theta$ as $\sin\theta^2$.**
$\sin^2\theta$ means $(\sin\theta)^2$, not $\sin(\theta^2)$. This notation confusion can lead to completely wrong calculations.

Mistake 3: Incorrect values at standard angles.
A very common error is writing $\sin 60^\circ = \dfrac{1}{2}$ (it should be $\dfrac{\sqrt{3}}{2}$) or $\cos 30^\circ = \dfrac{1}{2}$ (it should be $\dfrac{\sqrt{3}}{2}$). Use the memory trick: for $\sin$ values, the numerator follows $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$.

Mistake 4: Starting identity proofs from both sides simultaneously.
In CBSE exams, you should start from one side (typically the more complex one) and simplify it step by step until it equals the other side. Working from both sides simultaneously and meeting in the middle is not considered a valid proof.

**Mistake 5: Forgetting that $\tan 90^\circ$ and $\sec 90^\circ$ are undefined.**
$\cos 90^\circ = 0$, so $\tan 90^\circ = \sin 90^\circ / \cos 90^\circ$ and $\sec 90^\circ = 1/\cos 90^\circ$ are both undefined (division by zero). Similarly, $\csc 0^\circ$ and $\cot 0^\circ$ are undefined.

Q1. If $\cos\theta = \dfrac{5}{13}$, find $\sin\theta$ and $\tan\theta$.

Answer: Adjacent $= 5$, hypotenuse $= 13$, opposite $= \sqrt{169 - 25} = 12$. $\sin\theta = \dfrac{12}{13}$, $\tan\theta = \dfrac{12}{5}$.

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Q2. Evaluate: $\sin^2 30^\circ + \sin^2 45^\circ + \sin^2 60^\circ$.

Answer: $\dfrac{1}{4} + \dfrac{1}{2} + \dfrac{3}{4} = \dfrac{1 + 2 + 3}{4} = \dfrac{6}{4} = \dfrac{3}{2}$.

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Q3. Show that $\cos 38^\circ \cos 52^\circ - \sin 38^\circ \sin 52^\circ = 0$.

Answer: $\cos 52^\circ = \sin 38^\circ$ and $\sin 52^\circ = \cos 38^\circ$ (complementary). So the expression becomes $\cos 38^\circ \sin 38^\circ - \sin 38^\circ \cos 38^\circ = 0$.

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Q4. Prove: $\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \dfrac{1}{\sec\theta - \tan\theta}$.

Answer: Divide numerator and denominator by $\cos\theta$: $\dfrac{\tan\theta - 1 + \sec\theta}{\tan\theta + 1 - \sec\theta}$. Using $\sec^2\theta - \tan^2\theta = 1$, multiply and simplify to get $\sec\theta + \tan\theta = \dfrac{1}{\sec\theta - \tan\theta}$.

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Q5. If $\tan(A + B) = \sqrt{3}$ and $\tan(A - B) = \dfrac{1}{\sqrt{3}}$, where $0^\circ < A + B \leq 90^\circ$, find $A$ and $B$.

Answer: $A + B = 60^\circ$ and $A - B = 30^\circ$. Adding: $2A = 90^\circ \implies A = 45^\circ$. Subtracting: $2B = 30^\circ \implies B = 15^\circ$.

• The six trigonometric ratios ($\sin$, $\cos$, $\tan$, $\csc$, $\sec$, $\cot$) are defined for acute angles in a right triangle.
  - Given any one ratio, all other ratios can be found using the Pythagoras theorem.
  - The standard angle table ($0^\circ$ to $90^\circ$) must be memorised — it is used in nearly every problem.
  - Complementary angle relationships convert between $\sin$ and $\cos$, $\tan$ and $\cot$, $\sec$ and $\csc$.
  - The three identities ($\sin^2 + \cos^2 = 1$, $1 + \tan^2 = \sec^2$, $1 + \cot^2 = \csc^2$) are the foundation for all identity proofs.
  - For identity proofs: start from the complex side, convert to $\sin/\cos$ if stuck, and look for factoring opportunities.
  - This chapter combined with Chapter 9 typically carries $6$-$10$ marks in the board exam.

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