OP Malhotra Class 10 Maths Solutions Chapter 16 Trigonometrical Identities and Tables Ex 16(a)

Students often turn to Class 10 ICSE Maths Solutions S Chand Chapter 16 Trigonometrical Identities and Tables Ex 16(a) to clarify doubts and improve problem-solving skills.

S Chand Class 10 ICSE Maths Solutions Chapter 16 Trigonometrical Identities and Tables Ex 16(a)

Question 1.
\(\frac{1-\cos ^2 \theta}{\sin ^2 \theta}\) = 1.
Solution:
L.H.S. = \(\frac{1-\cos ^2 \theta}{\sin ^2 \theta}\) = \(\frac{\sin ^2 \theta}{\sin ^2 \theta}\) = 1 = R.H.S. {∵ 1 – cos² θ = sin² θ}

Question 2.
\(\frac{1-\sin ^2 \theta}{\cos ^2 \theta}\) = 1.
Solution:
L.H.S. = \(\frac{1-\sin ^2 \theta}{\cos ^2 \theta}\) = \(\frac{\cos ^2 \theta}{\cos ^2 \theta}\) = 1 = R.H.S. {∵ 1 – sin² θ = cos² θ}

Question 3.
sin A . cot A = cos A.
Solution:
L.H.S. = sin A . cot A = sin A = \(\frac{\cos A}{\sin A}\)

= cos A = R.H.S.

Question 4.
\(\frac{1}{\cos ^2 \theta}-\tan ^2 \theta=1\)
Solution:
L.H.S. = \(\frac{1}{\cos ^2 \theta}-\tan ^2 \theta\) = sec² θ – tan² θ

= 1 = R.H.S.

Question 5.
tan² A cos² A = 1 – cos² A.
Solution:
L.H.S. = tan² A cos² A = \(\frac{\sin ^2 A}{\cos ^2 A} \cos ^2 A\)

= sin² A = 1 – cos² A = R.H.S. {∵sin² A = 1 – cos² A}

Question 6.
tan θ = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)
Solution:
R.H.S. = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\) =\(\frac{\sin \theta}{\sqrt{\cos ^2 \theta}}\) {∵ 1 – sin² θ – cos² θ}
= \(\frac{\sin \theta}{\cos \theta}\) = tan θ = L.H.S.

Question 7.
\(\frac{1+\cos \theta}{\sin ^2 \theta}\) = \(\frac{1}{1-\cos \theta}\)
Solution:
L.H.S. = \(\frac{1+\cos \theta}{\sin ^2 \theta}\) = \(\frac{1+\cos \theta}{1-\cos ^2 \theta}\) {∵ sin² θ = 1 – cos² θ}
= \(\frac{1+\cos \theta}{(1+\cos \theta)(1-\cos \theta)}\) {∵ a² – b² = (a + b)(a – b)}
= \(\frac{1}{1-\cos \theta}\) = R.H.S.

Question 8.
cot² θ (1 – cos² θ) = cos² θ.
Solution:
L.H.S. = cot² θ (1 – cos² θ)
= \(\frac{\cos ^2 \theta}{\sin ^2 \theta} \times \sin ^2 \theta\)

= cos² θ = R.H.S.

Question 9.
tan² θ (1 – sin² θ) = sin² θ
Solution:
L.H.S. = tan²θ (1 – sin²θ) = \(\frac{\sin ^2 \theta}{\cos ^2 \theta} \times \cos ^2 \theta\)

= sin² θ = R.H.S.

Question 10.
(1 – sin² θ) sec² θ = 1.
Solution:
L.H.S. = (1 – sin² θ) sec² θ
= cos² θ × \(\frac{1}{\cos ^2 \theta}\)

= 1 = R.H.S.

Question 11.
(1 – cos² θ) cosec² θ = 1.
Solution:
L.H.S. = (1 – cos² θ) cosec² θ
= sin² θ × \(\frac{1}{\sin ^2 \theta}\)

= 1 = R.H.S.

Question 12.
sin² θ + \(\frac{1}{1+\tan ^2 \theta}\) = 1.
Solution:
L.H.S. = sin² θ + \(\frac{1}{1+\tan ^2 \theta}\) = sin² θ + \(\frac{1}{\sec ^2 \theta}\) {∵ 1 + tan² θ = sec² θ}
= sin² θ + cos² θ

= 1 = R.H.S. { ∵ sin² θ + cos² θ = 1}

Question 13.
\(\cos ^2 \theta+\frac{1}{1+\cot ^2 \theta}=1\)
Solution:
L.H.S. = cos² θ + \(\frac{1}{1+\cot ^2 \theta}\)

= cos² θ + sin² θ {∵sin² θ + cos² θ = 1}
= 1 = R.H.S.

Question 14.
\(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta}=1\)
Solution:
L.H.S. = \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta}\) = \(\frac { 1 }{ 1 }\)
{ ∵ sin² θ + cos² θ = 1 and sec² θ – tan² θ = 1}
= 1 = R.H.S.

Question 15.
\(\left(\frac{\cos ^2 A}{\sin ^2 A}+1\right)\) tan² A = \(\frac{1}{\cos ^2 A}\)
Solution:
L.H.S. = \(\left(\frac{\cos ^2 A}{\sin ^2 A}+1\right)\) tan² A
= (cot² A + 1) tan² A
= cosec² A × tan² A

= \(\frac{1}{\sin ^2 A}\) × \(\frac{\sin ^2 A}{\cos ^2 A}\) = \(\frac{1}{\cos ^2 A}\) = R.H.S.

Question 16.
sin⁴ θ + sin² θ cos² θ = sin² θ.
Solution:
L.H.S. = sin⁴ θ + sin² θ cos² θ
= sin² θ(sin² θ + cos² θ) {∵ sin² θ + cos² θ = 1}
= sin² θ × 1 = sin² θ = R.H.S.

Question 17.
sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ = 1.
Solution:
L.H.S. = sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ
= (sin² θ)² + 2 sin² θ cos² θ + (cos² θ)²
= (sin² θ + cos² θ)²

= (1)² = 1 = R.H.S.

Question 18.
sin⁴ A cosec² A + cos⁴ A sec² A = 1.
Solution:
L.H.S. = sin² A cosec² A + cos² A sec² A
= sin⁴ A × \(\frac{1}{\sin ^2 A}\) + cos⁴ A × \(\frac{1}{\cos ^2 A}\)

= sin² θ + cos² θ = 1 = R.H.S. {∵ sin² θ + cos² θ = 1}

Question 19.
sin² A cot² A + cos² A tan² A = 1.
Solution:
L.H.S. = sin² A cot² A + cos² A tan² A
= sin² A × \(\frac{\cos ^2 A}{\sin ^2 A}\) + cos² A × \(\frac{\sin ^2 A}{\cos ^2 A}\)

= cos² A + sin² A = 1 = R.H.S. {∵ sin² A + cos² A = 1}

Question 20.
tan θ + cot θ = sec θ cosec θ
Solution:
L.H.S. = tan θ + cot θ = \(\frac{\sin \theta}{\cos \theta}\) + \(\frac{\cos \theta}{\sin \theta}\)

= \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\) = \(\frac{1}{\cos \theta \sin \theta}\) = sec θ cosec θ

= R.H.S.

Question 21.
(tan A + cot A) sin A cos A = 1
Solution:
L.H.S. = (tan A + cot A) (sin A cos A)
= \(\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)\) sin A cos A

= \(\frac{\sin ^2 A+\cos ^2 A}{\cos A \sin A}\) × sin A cos A
= \(\frac{1}{\cos A \sin A}\) × sin A cos A { ∵ sin²θ + cos² θ = 1}
= 1 = R.H.S.

Question 22.
\(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)}=\cot \theta\)
Solution:
L.H.S. = \(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)}\)
= \(\frac{1-\sin ^2 \theta+\cos \theta}{\sin \theta(1+\cos \theta)}\) {∵ 1 – sin² θ = cos² θ}
= \(\frac{\cos \theta+\cos ^2 \theta}{\sin \theta(1+\cos \theta)}\) = \(\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}\) = \(\frac{\cos \theta}{\sin \theta}\) = cot θ = R.H.S.

Question 23.
\(\frac{1}{1-\cos \theta}\) + \(\frac{1}{1+\cos \theta}\) = 2 cosec² θ
Solution:
L.H.S. = \(\frac{1}{1-\cos \theta}\) + \(\frac{1}{1+\cos \theta}\) = \(\frac{1+\cos \theta+1-\cos \theta}{(1-\cos \theta)(1+\cos \theta)}\) = \(\frac{2}{1-\cos ^2 \theta}\) {(a + b) (a – b) = a² – b²}
= \(\frac{2}{\sin ^2 \theta}\) = 2 cosec² θ

=R.H.S.

Question 24.
\(\frac{1-\tan ^2 \theta}{\cot ^2 \theta-1}=\tan ^2 \theta\)
Solution:

Question 25.
\(\frac{1}{\sec \theta+\tan \theta}\) = \(\frac{1-\sin \theta}{\cos \theta}\)
Solution:
L.H.S. = \(\frac{1}{\sec \theta+\tan \theta}\)

Question 26.
(cosec A – sin A) (sec A – cos A) (tan A + cot A ) = 1.
Solution:

Question 27.
\(\frac{1+\sin \theta}{1-\sin \theta}\) = (sec θ + tan θ)²
Solution:
L.H.S. = \(\frac{1+\sin \theta}{1-\sin \theta}\) = \(\frac{(1+\sin \theta)(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}\) = \(\frac{(1+\sin \theta)^2}{1-\sin ^2 \theta}\) = \(\frac{(1+\sin \theta)^2}{\cos ^2 \theta}\) (∵ 1 – sin² θ = cos² θ)
= \(\left(\frac{1+\sin \theta}{\cos \theta}\right)^2\) = \(\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right)^2\)
= (sec θ + tan θ)² = R.H.S.

Question 28.
\(\frac{1+\cos \theta}{1-\cos \theta}\) = (cosec θ + cot θ)²
Solution:

Question 29.
= cosec θ + cot θ = \(\frac{1+\cos \theta}{\sin \theta}\)
Solution:

Question 30.
If tan θ + cot θ = 2, Prove that tan² θ + cot² θ = 2.
Solution:
tan θ + cot θ = 2
Squaring both sides,
(tan θ + cot θ)² = 4
tan² θ + cot² θ + 2 tan θ cot θ = 4
⇒ tan² θ + cot² θ + 2 × 1 = 4 { ∵ tan θ cot θ = 1}
⇒ tan² θ + cot² θ + 2 = 4
⇒ tan² θ + cot² θ = 4 – 2 = 2
∴ tan² θ + cot² θ = 2

Question 31.
If sin θ + cos θ = a, sin θ – cos θ = b, Prove that a² + b² = 2
Solution:
a² + b² = (sin θ + cos θ)² + (sin θ – cos θ)²
= sin² θ + cos² θ + 2 sin θ cos θ + sin² θ + cos² θ – 2sin θ cos θ
= 2 sin² θ + 2 cos² θ = 2 (sin² θ + cos² θ)
= 2 × 1 = 2 {∵ sin² θ + cos² θ = 1} Hence Proved.