CBSE Class 10 Mathematics Trigonometry Notes Set 16

Trigonometric Identities

\[ \text{(i)}\ \sin^2\theta + \cos^2\theta = 1 \quad (\text{so } \sin^2\theta = 1 - \cos^2\theta \text{ and } \cos^2\theta = 1 - \sin^2\theta) \]

\[ \text{(ii)}\ 1 + \tan^2\theta = \sec^2\theta \quad (\theta \ne 90^\circ) \quad (\text{so } \sec^2\theta - \tan^2\theta = 1) \]

\[ \text{(iii)}\ 1 + \cot^2\theta = \operatorname{cosec}^2\theta \quad (\theta \ne 0^\circ) \quad (\text{so } \operatorname{cosec}^2\theta - \cot^2\theta = 1) \]

Proof. Consider a right-angled \( \triangle ABC \) in which \( \angle B = 90^\circ \) and \( \angle A = \theta^\circ \). Let AB = x units, BC = y units and AC = r units.

Then, by the Pythagoras theorem, \( x^2 + y^2 = r^2 \). Now

\[ \text{(i)}\ \sin^2\theta + \cos^2\theta = \left(\frac{y}{r}\right)^2 + \left(\frac{x}{r}\right)^2 = \frac{x^2 + y^2}{r^2} = \frac{r^2}{r^2} = 1 \]

\[ \text{(ii)}\ 1 + \tan^2\theta = 1 + \left(\frac{y}{x}\right)^2 = \frac{x^2 + y^2}{x^2} = \left(\frac{r}{x}\right)^2 = \sec^2\theta \]

\[ \text{(iii)}\ 1 + \cot^2\theta = 1 + \left(\frac{x}{y}\right)^2 = \frac{x^2 + y^2}{y^2} = \left(\frac{r}{y}\right)^2 = \operatorname{cosec}^2\theta \]

Solved Examples

Example 11. Prove that \( \cot\theta - \tan\theta = \dfrac{2\cos^2\theta - 1}{\sin\theta\cos\theta} \).
Answer:
LHS \( = \dfrac{\cos\theta}{\sin\theta} - \dfrac{\sin\theta}{\cos\theta} = \dfrac{\cos^2\theta - \sin^2\theta}{\sin\theta\cos\theta} = \dfrac{\cos^2\theta - (1 - \cos^2\theta)}{\sin\theta\cos\theta} = \dfrac{2\cos^2\theta - 1}{\sin\theta\cos\theta} = \) RHS. Hence proved.

Example 12. Prove that \( (\operatorname{cosec}A - \sin A)(\sec A - \cos A)(\tan A + \cot A) = 1 \).
Answer:
LHS \( = \left(\dfrac{1}{\sin A} - \sin A\right)\left(\dfrac{1}{\cos A} - \cos A\right)\left(\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}\right) \)
\( = \dfrac{1 - \sin^2A}{\sin A} \cdot \dfrac{1 - \cos^2A}{\cos A} \cdot \dfrac{\sin^2A + \cos^2A}{\sin A\cos A} = \dfrac{\cos^2A}{\sin A} \cdot \dfrac{\sin^2A}{\cos A} \cdot \dfrac{1}{\sin A\cos A} = 1 = \) RHS. Hence proved.

Example 13. If \( \sin\theta + \cos\theta = m \) and \( \sec\theta + \operatorname{cosec}\theta = n \), then prove that \( n(m^2 - 1) = 2m \).
Answer:
LHS \( = (\sec\theta + \operatorname{cosec}\theta)[(\sin\theta + \cos\theta)^2 - 1] = \left(\dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta}\right)(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta - 1) \)
\( = \dfrac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} \cdot (2\sin\theta\cos\theta) = 2(\sin\theta + \cos\theta) = 2m = \) RHS. Hence proved.

Example 14. If \( \sec\theta = x + \dfrac{1}{4x} \), then prove that \( \sec\theta + \tan\theta = 2x \) or \( \dfrac{1}{2x} \).
Answer:
\( \tan^2\theta = \sec^2\theta - 1 = \left(x + \dfrac{1}{4x}\right)^2 - 1 = x^2 + \dfrac{1}{16x^2} + \dfrac{1}{2} - 1 = \left(x - \dfrac{1}{4x}\right)^2 \).
So \( \tan\theta = x - \dfrac{1}{4x} \) (ii) or \( \tan\theta = -\left(x - \dfrac{1}{4x}\right) \) (iii).
Adding (i) and (ii): \( \sec\theta + \tan\theta = x + \dfrac{1}{4x} + x - \dfrac{1}{4x} = 2x \).
Adding (i) and (iii): \( \sec\theta + \tan\theta = x + \dfrac{1}{4x} - x + \dfrac{1}{4x} = \dfrac{1}{2x} \).
Hence \( \sec\theta + \tan\theta = 2x \) or \( \dfrac{1}{2x} \).

Example 15. If \( \theta \) is an acute angle and \( \tan\theta + \cot\theta = 2 \), find the value of \( \tan^9\theta + \cot^9\theta \).
Answer:
\( \tan\theta + \dfrac{1}{\tan\theta} = 2 \Rightarrow \tan^2\theta - 2\tan\theta + 1 = 0 \Rightarrow (\tan\theta - 1)^2 = 0 \Rightarrow \tan\theta = 1 \Rightarrow \theta = 45^\circ \).
\( \tan^9 45^\circ + \cot^9 45^\circ = 1 + 1 = \) 2.

Example 16. Evaluate: \( \dfrac{\sec^2 54^\circ - \cot^2 36^\circ}{\operatorname{cosec}^2 57^\circ - \tan^2 33^\circ} + 2\sin^2 38^\circ\sec^2 52^\circ - \sin^2 45^\circ + \dfrac{2}{\sqrt{3}}\tan 17^\circ\tan 60^\circ\tan 73^\circ \).
Answer:
Using complementary angles: \( \sec 54^\circ = \operatorname{cosec}36^\circ \), \( \operatorname{cosec}57^\circ = \sec 33^\circ \), \( \sec 52^\circ = \operatorname{cosec}38^\circ \), \( \tan 73^\circ = \cot 17^\circ \).
The first term is \( \dfrac{\operatorname{cosec}^2 36^\circ - \cot^2 36^\circ}{\sec^2 33^\circ - \tan^2 33^\circ} = \dfrac{1}{1} = 1 \). The second term is \( 2\sin^2 38^\circ \cdot \dfrac{1}{\sin^2 38^\circ} = 2 \). The third term is \( -\dfrac{1}{2} \). The fourth term is \( \dfrac{2}{\sqrt{3}} \cdot \sqrt{3} \cdot (\tan 17^\circ\cot 17^\circ) = 2 \).
Total \( = 1 + 2 - \dfrac{1}{2} + 2 = \) 9/2.

Check Your Progress

Check Your Progress 1. Express \( \cos\theta \) in terms of \( \sin\theta \).
Answer:
\( \cos\theta = \sqrt{1 - \sin^2\theta} \) (for an acute angle).

Check Your Progress 2. Express \( \tan\theta \) in terms of \( \cos\theta \).
Answer:
\( \tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{\sqrt{1 - \cos^2\theta}}{\cos\theta} \) (for an acute angle).

Check Your Progress 3. Prove that \( (\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta \).
Answer:
LHS \( = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + 2\sin\theta\cos\theta \).

Check Your Progress 4. Prove that \( \cot^2\theta + \operatorname{cosec}^2\theta = \operatorname{cosec}^4\theta - \cot^4\theta \).
Answer:
RHS \( = (\operatorname{cosec}^2\theta - \cot^2\theta)(\operatorname{cosec}^2\theta + \cot^2\theta) = 1 \cdot (\operatorname{cosec}^2\theta + \cot^2\theta) = \) LHS.

Check Your Progress 5. Prove that \( \dfrac{\tan\theta}{\sec\theta - 1} + \dfrac{\tan\theta}{\sec\theta + 1} = 2\operatorname{cosec}\theta \).
Answer:
LHS \( = \tan\theta \cdot \dfrac{(\sec\theta + 1) + (\sec\theta - 1)}{\sec^2\theta - 1} = \tan\theta \cdot \dfrac{2\sec\theta}{\tan^2\theta} = \dfrac{2\sec\theta}{\tan\theta} = \dfrac{2}{\sin\theta} = 2\operatorname{cosec}\theta \).

Check Your Progress 6. If \( x = a\cos\theta \) and \( y = a\sin\theta \), what is the value of \( x^2 + y^2 \)?
Answer:
\( x^2 + y^2 = a^2(\cos^2\theta + \sin^2\theta) = \) a².

Check Your Progress 7. If \( x = \operatorname{cosec}\theta - \cot\theta \) and \( y = \operatorname{cosec}\theta + \cot\theta \), find xy.
Answer:
\( xy = \operatorname{cosec}^2\theta - \cot^2\theta = \) 1.