Question

The value of $\theta$ [0° < $\theta$ < 90°] for with $\sf {\dfrac {\cos \; \theta}{1-\sin \theta} + \dfrac {\cos \; \theta}{1+\sin \theta} = 4}$ is a) 45° b) 60° c) 30° d) None of these​

Answer 1

Given :

• $\sf {\dfrac {\cos \; \theta}{1-\sin \theta} + \dfrac {\cos \; \theta}{1+\sin \theta} = 4}$

To calculate :

• Value of $\theta$.

Solution :

According to the question,

$\twoheadrightarrow \sf {\dfrac {\cos \; \theta}{1-\sin \theta} + \dfrac {\cos \; \theta}{1+\sin \theta} = 4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac {\cos \; \theta(1 + \sin \; \theta) + cos \; \theta(1 - sin \; \theta)}{(1-\sin \theta)(1+\sin \theta)}= 4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac {\cos \; \theta(1 + \sin \; \theta + 1 -\sin \; \theta) }{(1-\sin \theta)(1+\sin \theta)}= 4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac {\cos \; \theta(1 + 1 ) }{(1-\sin \theta)(1+\sin \theta)}= 4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac {\cos \; \theta(2) }{1 - \sin^2\theta}= 4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac {2\cos \; \theta }{cos^2 \;\theta}= 4}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac {2 }{cos \; \theta}= 4}$

By cross multiplication,

$\twoheadrightarrow \sf {2(1)= 4cos\; \theta}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {2= 4cos \; \theta}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac{2}{4}= cos \; \theta}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {\dfrac{1}{2}= cos \; \theta}$

⠀⠀⠀⠀⠀⠀

• Cos 60° = 1/2 , So, ⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\twoheadrightarrow \sf {cos \;60^\circ= cos \; \theta}$

Therefore,

$\dashrightarrow \boxed{\bf {60^\circ= \theta}}$

Henceforth, the value of $\theta$ is 60°. Option B) is correct.

Answer 2

Question:-

$\sf The\: value\: of\: \theta \left[0° \lt \theta \lt 90°\right] \:for \:which, \\ \sf {\dfrac {\cos \; \theta}{1-\sin \theta} + \dfrac {\cos \; \theta}{1+\sin \theta} = 4}\:is$

$\bigcirc\sf 45°$

$\bigodot\bf 60°$

$\bigcirc\sf 30°$

$\bigcirc\sf NOTA$

Solution:-

$\\ \sf\longmapsto \dfrac{cos\Theta}{1-sin\Theta}+\dfrac{cos\Theta}{1+sin\Theta}=4$

$\\ \sf\longmapsto \dfrac{cos\Theta+cos\Theta sin\Theta +cos\Theta-cos\Theta sin\Theta}{(1-sin\Theta)(1+sin\Theta)}=4$

$\\ \sf\longmapsto \dfrac{cos\Theta+cos\Theta+cos\Theta sin\Theta-cos\Theta sin\Theta}{(1)^2-(sin\Theta)^2}=4$

$\\ \sf\longmapsto \dfrac{2cos\Theta}{1-sin^2\Theta}=4$

We know

$\boxed{\sf 1-sin^2\Theta=cos^2\Theta}$

$\\ \sf\longmapsto \dfrac{\cancel{2cos\Theta}}{\cancel{cos^2\Theta}}=4$

$\\ \sf\longmapsto \dfrac{2}{cos\Theta}=4$

• Cancel 2 from both sides

$\\ \sf\longmapsto \dfrac{1}{cos\Theta}=2$

$\\ \sf\longmapsto 2cos\Theta=1$

$\\ \sf\longmapsto cos\Theta=\dfrac{1}{2}$

$\boxed{\sf cos60°=\dfrac{1}{2}}$

$\\ \sf\longmapsto cos\Theta=cos60°$

• Cancelling cos

$\\ \sf\longmapsto \Theta=60°$

Option b is correct