The value of tan^ - 1 [ 1 - sin x + 1 + sin x 1 - sin x - 1 + sin x ] ( x ∈ [0, (π)/(2)]) is equal to

Correct answer: A. (x)/(2) - (π)/(2) ta n ^ - 1 [ 1 - sin x + 1 + sin x 1 - sin x - 1 + sin x ] = ta n ^ - 1 [ ( 1 - sin x + 1 + sin x ) ( 1 - sin x - 1 + sin x ) ( 1 - sin x + 1 + sin x ) ( 1 - sin x + 1 + sin x ) ] = ta n ^ - 1 [ (1 - sin x) + (1 + sin x) + 2 1 - si n ^ 2 x (1 - sin x) - (1 + sin x) ] = ta n ^ - 1 [(2(1 + cos x))/(- 2sin x)] = ta n ^ - 1 [ - 2co s ^ 2 ((x)/(2)) 2 ((x)/(2)) ((x)/(2)) ] = ta n ^ - 1 (- cot(x)/(2)) = ta n ^ - 1 [ (π - (x)/(2))] = (π)/(2) - co t ^ - 1 [ (π - (x)/(2))] = (π)/(2) - (π - (x)/(2)) = (x)/(2) - (π)/(2)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

The value of tan^ - 1 [ 1 - sin x + 1 + sin x 1 - sin x - 1 + sin x...

The value of

tan^{- 1} \left[\dfrac{\sqrt{1 - sin x} + \sqrt{1 + sin x}}{\sqrt{1 - sin x} - \sqrt{1 + sin x}}\right] \left(\forall x \in \left[0, \dfrac{\pi}{2}\right]\right)

is equal to

Options

A $\dfrac{x}{2} - \dfrac{\pi}{2}$
B $\dfrac{x}{2} + \dfrac{\pi}{2}$
C $\dfrac{x}{2} - \pi$
D $\dfrac{\pi}{2} - \dfrac{x}{2}$

Correct answer

A. $\dfrac{x}{2} - \dfrac{\pi}{2}$

Step-by-step solution

ta{n}^{- 1}\left[\dfrac{\sqrt{1 - sin x} + \sqrt{1 + sin x}}{\sqrt{1 - sin x} - \sqrt{1 + sin x}}\right]

= ta{n}^{- 1}\left[\dfrac{\left(\sqrt{1 - sin x} + \sqrt{1 + sin x}\right)}{\left(\sqrt{1 - sin x} - \sqrt{1 + sin x}\right)}\dfrac{\left(\sqrt{1 - sin x} + \sqrt{1 + sin x}\right)}{\left(\sqrt{1 - sin x} + \sqrt{1 + sin x}\right)}\right]

= ta{n}^{- 1}\left[\dfrac{\left(1 - sin x\right) + \left(1 + sin x\right) + 2\sqrt{1 - si{n}^{2}x}}{\left(1 - sin x\right) - \left(1 + sin x\right)}\right] = ta{n}^{- 1}\left[\dfrac{2\left(1 + cos x\right)}{- 2sin x}\right]

= ta{n}^{- 1}\left[\dfrac{- 2co{s}^{2}\left(\dfrac{x}{2}\right)}{2\sin \left(\dfrac{x}{2}\right)\cos \left(\dfrac{x}{2}\right)}\right] = ta{n}^{- 1}\left(- cot\dfrac{x}{2}\right) = ta{n}^{- 1}\left[\cot \left(\pi - \dfrac{x}{2}\right)\right]

= \dfrac{\pi}{2} - co{t}^{- 1}\left[\cot \left(\pi - \dfrac{x}{2}\right)\right] = \dfrac{\pi}{2} - \left(\pi - \dfrac{x}{2}\right) = \dfrac{x}{2} - \dfrac{\pi}{2}

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