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

Parents can use S Chand Class 11 Maths Solutions Chapter 4 Trigonometrical Functions Ex 4(a) to provide additional support to their children.

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

Prove that

Question 1.
1 – cos² θ = sin² θ.
Solution:
L.H.S. = 1 – cos² θ = 1 – \(\frac { x² }{ r² }\)
= \(\frac{r^2-x^2}{r^2}=\frac{x^2+y^2-x^2}{r^2}\)
= \(\frac{y^2}{r^2}\) = sin² θ = R.H.S

Question 2.
\(\sqrt{1-\sin ^2 \theta}\) = cos θ
Solution:
L.H.S = \(\sqrt{1-\sin ^2 \theta}\)
= \(\sqrt{1-\frac{y^2}{r^2}}=\sqrt{\frac{r^2-y^2}{r^2}}=\sqrt{\frac{x^2}{r^2}}\)
= ± \(\frac { x }{ r }\) = ± cos θ [∵ x = r cos θ]

Question 3.
Sec α \(\sqrt{1-\sin ^2 \alpha}\) = 1
Solution:
L.H.S = sec α \(\sqrt{1-\sin ^2 \alpha}\)
= sec α \(\sqrt{\cos ^2 \alpha}\) = sec α | cos α |
= sec α x cos α = 1 if a lies in first and IVth quadrant. [in this case cos α > θ]

Question 4.
\(\frac{\sec ^2 \theta-1}{\tan ^2 \theta}\) = 1
Solution:

Question 5.
sec² θ – 1 – tan² θ = 0.
Solution:
L.H.S = sec² θ – 1 – tan² θ
= \(\frac{r^2}{x^2}-1-\frac{y^2}{x^2}=\frac{r^2-x^2-y^2}{x^2}\)
= \(\frac{r^2-r^2}{x^2}\) = 0 = R.H.S.

Question 6.
\(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta}\) = 1.
Solution:
L.H.S = \(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sec ^2 \theta-\tan ^2 \theta}\)
= \(\frac{\frac{y^2}{r^2}+\frac{x^2}{r^2}}{\frac{r^2}{x^2}-\frac{y^2}{x^2}}=\frac{\frac{x^2+y^2}{r^2}}{\frac{r^2-y^2}{x^2}}=\frac{\frac{r^2}{r^2}}{\frac{x^2}{x^2}}\) = 1
= R.H.S [∵ r² = x² + y²]

Question 7.
1 – cos² θ – sin² θ = 0.
Solution:
L.H.S = 1 – cos² θ – sin² θ
= 1 – (cos² θ + sin² θ) = 1 – 1 = 0
= R.H.S
Aliter : L.H.S = 1 – cos² θ – sin² θ
= 1 – \(\frac{x^2}{r^2}-\frac{y^2}{r^2}\)
= \(\frac{r^2-x^2-y^2}{r^2}=\frac{r^2-r^2}{r^2}\) = 0

Question 8.
sin⁴ θ + sin² θ cos² θ = sin² 0.
Solution:
L.H.S = sin⁴ θ + sin² θ cos² θ
= sin² θ (sin² θ + cos² θ)
= sin² θ = R.H.S.
Aliter: L.H.S = \(\frac{y^4}{r^4}+\frac{y^2}{r^2} \cdot \frac{x^2}{r^2}\)
= \(\frac{y^2\left(x^2+y^2\right)}{r^4}\)
= \(\frac{y^2 \cdot r^2}{r^4}=\frac{y^2}{r^2}\)
= sin² θ = R.H.S

Question 9.
sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ = 1.
Solution:
L.H.S = sin⁴ θ + 2 sin² θ cos² θ + cos² θ
= (sin² θ + cos² θ)² = 1² = 1
= R.H.S.
Aliter:
L.H.S = sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ
= \(\frac{y^4}{r^2}+\frac{2 y^2}{r^2} \cdot \frac{x^2}{r^2}+\frac{x^4}{r^4}=\frac{\left(y^2+x^2\right)^2}{r^4}\)
= \(\frac{r^4}{r^4}\) = 1 = R.H.S

Question 10.
cos² θ (cosec² θ – cot² θ) = cot² θ.
Solution:
L.H.S = cos² θ (cosec² θ – cot² θ)
= cos² θ . 1 = cos² θ = R.H.S [∵ 1 + cot² θ = cosec² θ]
Aliter:
L.H.S = cos² θ (cosec² θ – cot² θ)
= \(\frac{x^2}{r^2}\left[\frac{r^2}{y^2}-\frac{x^2}{y^2}\right]=\frac{x^2}{r^2}\left[\frac{r^2-x^2}{y^2}\right]\)
= \(\frac{x^2}{r^2}\left[\frac{x^2+y^2-x^2}{y^2}\right]\)
= \(\frac{x^2}{r^2}\) = cos² θ = R.H.S

Question 11.
\(\frac{\sin \theta {cosec} \theta \tan \theta \cot \theta}{\sin ^2 \theta+\cos ^2 \theta}\) = 1.
Solution:

Question 12.
\(\frac{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}{\sec ^2 57^{\circ}-\tan ^2 57^{\circ}}\) = 1
Solution:
L.H.S = \(\frac{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}{\sec ^2 57^{\circ}-\tan ^2 57^{\circ}}\)
= \(\frac { 1 }{ 1 }\) = 1
= R.H.S.
[∵ sin² 9 + cos² 9=1 = sec² θ – tan² θ]

Question 13.
cos A tan A = sin A.
Solution:
L.H.S. = cos A tan A = \(\frac { x }{ r }\) x \(\frac { y }{ x }\)
= \(\frac { y }{ r }\) = sin A
= R.H.S.

Question 14.
sin⁴ A cosec² A + cos⁴ A sec² A = 1.
Solution:

Question 15.
sin² A cot² A + cos² A tan² A = 1.
Solution:
L.H.S = sin² A cot² A + cos² A tan² A
= \(\frac{y^2}{r^2} \times \frac{x^2}{y^2}+\frac{x^2}{r^2} \times \frac{y^2}{x^2}\)
= \(\frac{x^2}{r^2}+\frac{y^2}{r^2}=\frac{x^2+y^2}{r^2}=\frac{r^2}{r^2}\) = 1
= R.H.S.
Aliter:
L.H.S = sin² A cot² A + cos² A tan² A
= cos² A + sin² A = 1
= R.H.S