Trigonometric Identities Class 10 – Solved Problems & Examples

ANAND CLASSES Study Material and Notes to learn Trigonometric Identities for Class 10 with solved problems, step-by-step explanations, and NCERT-based questions to boost your board exam preparation.

Prove the following Trigonometric Identities:

1. (1 – cos² A) cosec² A = 1

Solution:

Taking the L.H.S.,

(1 – cos² A) cosec² A

= (sin² A) cosec² A [∵ sin² A + cos² A = 1 ⇒1 – sin² A = cos² A]

= 1²

= 1 = R.H.S.

– Hence, proved.

2. (1 + cot² A) sin² A = 1 

Solution: 

By using the identity,

cosec² A – cot² A = 1 ⇒ cosec² A = cot² A + 1

Taking,

L.H.S. = (1 + cot² A) sin² A

= cosec² A sin² A

= (cosec A sin A)²

= ((1/sin A) × sin A)²

= (1)²

= 1

= R.H.S.

– Hence, proved.

3. tan² θ cos² θ = 1 − cos² θ 

Solution: 

We know that,

sin² θ + cos² θ = 1

Taking,

L.H.S. = tan² θ cos² θ

= (tan θ × cos θ)²

= (sin θ)²

= sin² θ

= 1 – cos² θ

= R.H.S.

– Hence, proved.

4. cosec θ √(1 – cos² θ) = 1

Solution:

Using identity,

sin² θ + cos² θ = 1  ⇒ sin² θ = 1 – cos² θ

Taking L.H.S.,

L.H.S = cosec θ √(1 – cos² θ)

= cosec θ √( sin² θ)

= cosec θ x sin θ

= 1

= R.H.S.

– Hence, proved.

5. (sec² θ − 1)(cosec² θ − 1) = 1 

Solution:

Using identities,

(sec² θ − tan² θ) = 1 and (cosec² θ − cot² θ) = 1

We have,

L.H.S. = (sec² θ – 1)(cosec²θ – 1)

= tan²θ × cot²θ

= (tan θ × cot θ)²

= (tan θ × 1/tan θ)²

= 1²

= 1

= R.H.S.

– Hence, proved.

6. tan θ + 1/ tan θ = sec θ cosec θ

Solution:

We have,

L.H.S. = tan θ + 1/ tan θ

= (tan² θ + 1)/ tan θ

= sec² θ / tan θ [∵ sec² θ − tan² θ = 1]

= (1/cos² θ) x 1/ (sin θ/cos θ) [∵ tan θ = sin θ / cos θ]

= cos θ/ (sin θ x cos² θ)

= 1/ cos θ x 1/ sin θ

= sec θ x cosec θ

= sec θ cosec θ

= R.H.S.

– Hence, proved.

7. cos θ/ (1 – sin θ) = (1 + sin θ)/ cos θ

Solution:

We know that,

sin² θ + cos² θ = 1

So, by multiplying both the numerator and the denominator by (1+ sin θ), we get

$$ \frac{\cos \theta}{1 – \sin \theta}
= \frac{\cos \theta (1 + \sin \theta)}{(1 – \sin \theta)(1 + \sin \theta)}
$$
$$
= \frac{\cos \theta(1 + \sin \theta)}{(1 – \sin^2 \theta)}
$$
$$
= \frac{\cos \theta (1 + \sin \theta)}{\cos^2 \theta}
$$
$$
= \frac{1 + \sin \theta}{\cos \theta}
$$

$$\frac{\cos \theta}{1 – \sin \theta}= \frac{1 + \sin \theta}{\cos \theta}$$

L.H.S. = R.H.S.

– Hence, proved.

8. cos θ/ (1 + sin θ) = (1 – sin θ)/ cos θ

Solution:

We know that,

sin² θ + cos² θ = 1

So, by multiplying both the numerator and the denominator by (1- sin θ), we get

$$
\frac{\cos \theta}{1 + \sin \theta}
= \frac{\cos \theta (1 – \sin \theta)}{(1 + \sin \theta)(1 – \sin \theta)}
$$
$$
= \frac{\cos \theta (1 – \sin \theta)}{1 – \sin^2 \theta}
$$
$$
= \frac{\cos \theta (1 – \sin \theta)}{\cos^2 \theta}
$$
$$
= \frac{(1 – \sin \theta)}{\cos \theta}
$$
$$
\frac{\cos \theta}{1 + \sin \theta}
= \frac{1 – \sin \theta}{\cos \theta}
$$

L.H.S. = R.H.S.

– Hence, proved.

9. cos² A + 1/(1 + cot² A) = 1

Solution:

We already know that,

cosec² A − cot² A = 1 and sin² A + cos² A = 1

Taking L.H.S.,

$$
\text{L.H.S} = \cos^2 A + \frac{1}{1 + \cot^2 A}
$$
$$
= \cos^2 A + \frac{1}{\csc^2 A}
$$
$$
= \cos^2 A + \left( \frac{1}{\csc A} \right)^2
$$

= cos² A + sin² A = 1 = R.H.S.

– Hence, proved.

10. sin² A + 1/(1 + tan ² A) = 1

Solution:

We already know that,

sec² θ − tan² θ = 1 and sin² θ + cos² θ = 1

Taking L.H.S.,

$$
\text{L.H.S} = \sin^2 A + \frac{1}{1 + \tan^2 A}
$$
$$
= \sin^2 A + \frac{1}{\sec^2 A}
$$
$$
= \sin^2 A + \left( \frac{1}{\sec A} \right)^2
$$

= sin² A + cos² A = 1 = R.H.S.

– Hence, proved.

11.

$$\sqrt{\frac{1 – \cos \theta}{1 + \cos \theta}} $$

= cosec θ – cot θ

Solution:

We know that,

sin² θ + cos² θ = 1

Taking the L.H.S.,

$$
\text{L.H.S} = \sqrt{\frac{1 – \cos \theta}{1 + \cos \theta}}
$$
$$
= \sqrt{\frac{(1 – \cos \theta)(1 – \cos \theta)}{(1 + \cos \theta)(1 – \cos \theta)}}
$$
$$
= \sqrt{\frac{(1 – \cos \theta)^2}{1 – \cos^2 \theta}}
$$
$$
= \sqrt{\frac{(1 – \cos \theta)^2}{\sin^2 \theta}}
$$
$$
= \frac{1 – \cos \theta}{\sin \theta}
$$
$$
= \frac{1}{\sin \theta} – \frac{\cos \theta}{\sin \theta}
$$

= cosec θ – cot θ = R.H.S.

– Hence, proved.

12. 1 – cos θ/ sin θ = sin θ/ 1 + cos θ

Solution:

We know that,

sin² θ + cos² θ = 1

So, by multiplying both the numerator and the denominator by (1+ cos θ), we get

$$
\text{L.H.S} = \frac{1 – \cos^2 \theta}{(1 + \cos \theta)(\sin \theta)}
$$
$$
= \frac{\sin^2 \theta}{(1 + \cos \theta)(\sin \theta)}
$$
$$
= \frac{\sin \theta}{1 + \cos \theta}
$$

= R.H.S.

– Hence, proved.

13. sin θ/ (1 – cos θ) = cosec θ + cot θ

Solution:

Taking L.H.S.,

$$
\text{L.H.S} = \frac{\sin \theta}{1 – \cos \theta}
$$

On multiplying by its conjugate, we have

$$
= \frac{\sin \theta}{1 – \cos \theta} \times \frac{1 + \cos \theta}{1 + \cos \theta}
$$

$$
= \frac{\sin \theta (1 + \cos \theta)}{1 – \cos^2 \theta}
$$

$$
= \frac{\sin \theta + (\sin \theta \times \cos \theta)}{\sin^2 \theta}
$$

$$
= \frac{\sin \theta}{\sin^2 \theta} + \frac{\sin \theta \times \cos \theta}{\sin^2 \theta}
$$

$$
= \frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}
$$

= cosec θ + cot θ = R.H.S.

– Hence, proved.

14. (1 – sin θ) / (1 + sin θ) = (sec θ – tan θ)²

Solution:

Taking the L.H.S.,

$$
\text{L.H.S} = \frac{1 – \sin \theta}{1 + \sin \theta}
$$

On multiplying by its conjugate, we have

$$
= \frac{1 – \sin \theta}{1 + \sin \theta} \times \frac{1 – \sin \theta}{1 – \sin \theta}
$$

$$
= \frac{(1 – \sin \theta)(1 – \sin \theta)}{1 – \sin^2 \theta}
$$

$$
= \frac{(1 – \sin \theta)^2}{\cos^2 \theta}
$$

$$
= \left( \frac{1 – \sin \theta}{\cos \theta} \right)^2
$$

$$
= \left( \frac{1}{\cos \theta} – \frac{\sin \theta}{\cos \theta} \right)^2
$$

= (sec θ – tan θ)² = R.H.S.

– Hence, proved.

15. tan² θ − sin² θ = tan² θ sin² θ 

Solution:

Taking L.H.S.,

L.H.S = tan² θ − sin² θ 

$$
= \frac{\sin^2 \theta}{\cos^2 \theta} – \sin^2 \theta
$$

$$
= \sin^2 \theta \left( \frac{1}{\cos^2 \theta} – 1 \right)
$$

$$
= \sin^2 \theta \left( \frac{1 – \cos^2 \theta}{\cos^2 \theta} \right)
$$

$$
= \sin^2 \theta \left( \frac{\sin^2 \theta}{\cos^2 \theta} \right)
$$

= sin² θ tan² θ = R.H.S.

– Hence, proved.

16. (cosec θ + sin θ)(cosec θ – sin θ) = cot²θ + cos²θ 

Solution:

Taking L.H.S. = (cosec θ + sin θ)(cosec θ – sin θ)

On multiplying, we get

= cosec² θ – sin² θ

= (1 + cot² θ) – (1 – cos² θ) [Using cosec² θ − cot² θ = 1 and sin² θ + cos² θ = 1]

= 1 + cot² θ – 1 + cos² θ

= cot² θ + cos² θ

= R.H.S.

– Hence, proved.

17. (sec θ + cos θ) (sec θ – cos θ) = tan² θ + sin² θ 

Solution:

Taking L.H.S. = (sec θ + cos θ)(sec θ – cos θ)

On multiplying, we get,

= sec² θ – cos² θ

= (1 + tan² θ) – (1 – sin² θ) [Using sec² θ − tan² θ = 1 and sin² θ + cos² θ = 1]

= 1 + tan² θ – 1 + sin² θ

= tan ² θ + sin ² θ

= R.H.S.

– Hence, proved.

18. sec A(1- sin A) (sec A + tan A) = 1

Solution:

Taking L.H.S. = sec A(1 – sin A)(sec A + tan A)

Substituting sec A = 1/cos A and tan A =sin A/cos A in the above, we have,

L.H.S = 1/cos A (1 – sin A)(1/cos A + sin A/cos A)

= 1 – sin² A / cos² A [After taking L.C.M]

= cos² A / cos² A [∵ 1 – sin² A = cos² A]

= 1

= R.H.S.

– Hence, proved.

19. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1 

Solution:

Taking L.H.S. = (cosec A – sin A)(sec A – cos A)(tan A + cot A)

$$
= \left( \frac{1}{\sin A} – \sin A \right) \left( \frac{1}{\cos A} – \cos A \right) \left( \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} \right)
$$

$$
= \left( \frac{1 – \sin^2 A}{\sin A} \right) \left( \frac{1 – \cos^2 A}{\cos A} \right) \left( \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} \right)
$$

= (cos² A/ sin A) (sin² A/ cos A) (1/ sin A cos A) [∵ sin² θ + cos² θ = 1]

= (sin A  cos A)  (1/ cos A sin A)

= 1

= R.H.S.

– Hence, proved.

20. (1 + tan² θ)(1 – sin θ)(1 + sin θ) = 1

Solution:

Taking L.H.S. = (1 + tan²θ)(1 – sin θ)(1 + sin θ)

And, we know sin² θ + cos² θ = 1 and sec² θ – tan² θ = 1

So,

L.H.S = (1 + tan² θ)(1 – sin θ)(1 + sin θ)

= (1 + tan² θ){(1 – sin θ)(1 + sin θ)}

= (1 + tan² θ)(1 – sin² θ)

= sec² θ (cos² θ)

= (1/ cos² θ) x cos² θ

= 1

= R.H.S.

– Hence, proved.

21. sin² A cot² A + cos² A tan² A = 1

Solution:

We know that,

cot² A = cos² A/ sin² A and tan² A = sin² A/cos² A

Substituting the above in L.H.S., we get

L.H.S = sin² A cot² A + cos² A tan² A

= {sin² A (cos² A/ sin² A)} + {cos² A (sin² A/cos² A)}

= cos² A + sin² A

= 1 [∵ sin² θ + cos² θ = 1]

= R.H.S.

– Hence, proved.

22. (cos² θ/ sin θ) – cosec θ + sin θ = 0

Solution:

Taking L.H.S. and using sin² θ + cos² θ = 1, we have

$$
\text{L.H.S} = \frac{\cos^2 \theta}{\sin \theta} – \csc \theta + \sin \theta
$$

$$
= \left( \frac{\cos^2 \theta}{\sin \theta} – \csc \theta \right) + \sin \theta
$$

$$
= \left( \frac{\cos^2 \theta}{\sin \theta} – \frac{1}{\sin \theta} \right) + \sin \theta
$$

$$
= \left( \frac{\cos^2 \theta – 1}{\sin \theta} \right) + \sin \theta
$$

$$
= \left( \frac{-\sin^2 \theta}{\sin \theta} \right) + \sin \theta
$$

= – sin θ + sin θ = 0 = R.H.S.

– Hence, proved.

23. sec⁶ θ = tan⁶ θ + 3 tan² θ sec² θ + 1

Solution: 

From trig. identities we have,

sec² θ − tan² θ = 1

On cubing both sides,

(sec²θ − tan²θ)³ = 1

sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ(sec² θ − tan² θ) = 1[Since, (a – b)³ = a³ – b³ – 3ab(a – b)]

sec⁶ θ − tan⁶ θ − 3sec² θ tan² θ = 1

⇒ sec⁶ θ = tan⁶ θ + 3sec² θ tan² θ + 1

L.H.S. = R.H.S.

Hence, proved.

24. cosec⁶ θ = cot⁶ θ + 3cot² θ cosec² θ + 1

Solution:

From trig. identities we have,

cosec² θ − cot² θ = 1

On cubing both sides,

(cosec² θ − cot² θ)³ = 1

cosec⁶ θ − cot⁶ θ − 3cosec² θ cot² θ (cosec² θ − cot² θ) = 1[Since, (a – b)³ = a³ – b³ – 3ab(a – b)]

cosec⁶ θ − cot⁶ θ − 3cosec² θ cot² θ = 1

⇒ cosec⁶ θ = cot⁶ θ + 3 cosec² θ cot² θ + 1

L.H.S. = R.H.S.

Hence, proved.

25. sin⁴ A – cos⁴ A = 2 sin² A – 1

Solution:

Taking L.H.S,

sin⁴ A – cos⁴ A

= (sin² A)² – (cos² A)²

= (sin² A + cos² A) (sin² A – cos² A)

= sin²A – cos²A

= sin²A – (1 – sin²A) [Since, cos² A = 1 – sin² A]

= 2sin² A – 1

– Hence Proved

26. (1 – tan A)² + (1 + tan A)² = 2 sec² A

Solution:

Taking L.H.S,

(1 – tan A)² + (1 + tan A)²

= (1 + tan² A + 2 tan A) + (1 + tan² A – 2 tan A)

= 2 (1 + tan² A)

= 2 sec² A [Since, 1 + tan² A = sec² A]

– Hence Proved

27. cosec⁴ A – cosec² A = cot⁴ A + cot² A

Solution:

cosec⁴ A – cosec² A

= cosec² A(cosec² A – 1)

= (1 + cot² A) (1 + cot² A – 1)

= (1 + cot² A) cot² A

= cot⁴ A + cot² A = R.H.S

– Hence Proved

💡 Do You Know?

• The values of trigonometric ratios are exact for 0°, 30°, 45°, 60°, and 90°.
• These ratios are frequently used in geometry, physics, astronomy, and engineering.
• The identities like sin²θ + cos²θ = 1 help derive all other formulas in trigonometry.

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