OP Malhotra Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(b)

Students can track their progress and improvement through regular use of S Chand Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(b).

S Chand Class 11 ICSE Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(b)

Prove that

Question 1.
sec θ cot θ = cosec θ
Solution:
L.H.S = sec θ cot θ
= \(\frac{1}{\cos \theta} \frac{\cos \theta}{\sin \theta}=\frac{1}{\sin \theta}\)
= cosec θ

Question 2.
tan θ + cot θ = sec θ cosec θ
Solution:
L.H.S = tan θ + cot θ = \(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\)
= \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta}\)
= \(\frac{1}{\sin \theta \cos \theta}\)
= sec θ cosec θ = R.H.S.

Question 3.
\(\frac{\cos \theta}{\sec \theta-\tan \theta}\) = 1 + sin θ.
Solution:
L.H.S = \(\frac{\cos \theta}{\sec \theta-\tan \theta}\)
= \(\frac{\cos \theta}{\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}=\frac{\cos ^2 \theta}{1-\sin \theta}\)
= \(\frac{1-\sin ^2 \theta}{1-\sin \theta}\) = 1 + sin θ = R.H.S

Question 4.
\(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)}\) = cot θ.
Solution:
L.H.S. = \(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)}\)
= \(\frac{\cos ^2 \theta+\cos \theta}{\sin \theta(1+\cos \theta)}\)
= \(\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}\) = cot θ
= R.H.S

Question 5.
\(\frac{3-4 \sin ^2 \theta}{\cos ^2 \theta}\) = 3 – tan² θ.
Solution:
L.H.S = \(\frac{3-4 \sin ^2 \theta}{\cos ^2 \theta}\)
= 3 sec² θ – 4 tan² θ
= 3(1 + tan² θ) – 4 tan² θ
= 3 – tan² θ = R.H.S

Question 6.
(tan A + cot A) sin A x cos A = 1.
Solution:
L.H.S = (tan A + cot A) sin A cos A
= \(\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right) \sin A \cos A\)
= \(\left[\frac{\sin ^2 A+\cos ^2 A}{\sin A \cos A}\right] \sin A \cos A\)
= 1 = R.H.S.

Question 7.
cos A = \(\frac{\cot A}{{cosec} A}=\frac{\cot A}{\sqrt{\left(1+\cot ^2 A\right)}}\) (A ∉ III or IV quad.)
Solution:

Question 8.
sin⁴ θ – cos⁴ θ = sin² θ – cos² θ.
Solution:
L.H.S = sin⁴ θ – cos⁴ θ
= (sin² θ – cos² θ) (sin² θ + cos² θ)
= (sin² θ – cos2 θ) x 1
= sin² θ – cos² θ
= R.H.S

Question 9.
sec² A + cosec² A = sec² A cosec² A.
Solution:
L.H.S = sec² A + cosec² A
= \(\frac{1}{\cos ^2 A}+\frac{1}{\sin ^2 A}=\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 A \cos ^2 A}\)
= \(\frac{1}{\sin ^2 A \cos ^2 A}\) = sin² A cos² A
= R.H.S

Question 10.
sin³ θ + cos³ θ = (sin θ + cos θ) (1 – sin θ cos θ).
Solution:
sin³ θ + cos³ θ = (sin θ + cos θ) [sin² θ + cos² θ – sin θ cos θ]
= (sin θ + cos θ) (1 – sin θ cos θ) [∵ a³ + b³ = (a + b) (a² – ab + b²)]

Question 11.
sin² Φ (1 + cot² Φ) = 1.
Solution:
L.H.S = sin² Φ (1 + cot² Φ)
= sin² Φ cosec² Φ = 1
= R.H.S.

Question 12.
(tan A + cot A)² = cosec² A + sec² A.
Solution:
L.H.S = (tan A + cot A)²
= tan² A + cot² A + 2 tan A cot A
= sec² A – 1 + cosec² A – 1 + 2
= sec² A + cosec² A
= R.H.S.

Question 13.
sec⁴ A – sec² A = tan² A + tan4 A.
Solution:
L.H.S = sec⁴ A – sec² A
= sec² A (sec² A – 1)
= sec² A tan² A
= tan² A (1 + tan² A)
= R.H.S.

Question 14.
(cosec θ – sin θ) (sec θ – cos θ) (tan θ + cot θ) = 1.
Solution:

Question 15.
\(\frac{\cot A+\tan B}{\cot B+\tan A}\) = cot A tan B
Solution:

Question 16.
\(\frac{\sin \alpha}{1+\cos \alpha}+\frac{1+\cos \alpha}{\sin \alpha}\) = 2 cosec α.
Solution:

Question 17.
1 + \(\frac{1}{\cos A}=\frac{\tan ^2 A}{\sec A-1}\)
Solution:
R.H.S = \(\frac{\tan ^2 \mathrm{~A}}{\sec \mathrm{A}-1}\)
= \(\frac{\sec ^2 A-1}{\sec A-1}=\frac{(\sec A-1)(\sec A+1)}{\sec A-1}\)
= Sec A + 1 = \(\frac { 1 }{ cos A }\) + 1 = L.H.S

Question 18.
\(\frac{1+\cos A}{1-\cos A}\) = (cosec A + cot A)²
Solution:

Question 19.
If tan θ + sin θ = a and tan θ – sin θ = ß, show that α² – ß² = 4\(\sqrt{\alpha \beta}\). (θ ∉ II or III quad.)
Solution:
Given tan θ + sin θ = α … (1)
and tan θ – sin θ = ß … (2)

[∵ θ lies in either first or IVth quadrant ∴ cos θ > 0 and sin² θ > 0]
from (3) and (4) ; we have
L.H.S = R.H.S