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Question

$$\tan^{-1} \left[ \dfrac{\sqrt{1 + \sin x} + \sqrt{1 - \sin x}}{\sqrt{1 + \sin x} - \sqrt{1 - \sin x} }\right] ; \ \dfrac{dy}{dx} = ?$$

Solution
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$${\textbf{Step -1: Simplifying the given expression by applying relevant Trigonometric Identity}}{\textbf{.}}$$
$${\text{y = }}{\tan ^{ - 1}}\left[ {\dfrac{{\sqrt {1 + \sin {\text{x}}} + \sqrt {1 - \sin {\text{x}}} }}{{\sqrt {1 + \sin {\text{x}}} - \sqrt {1 - \sin {\text{x}}} }}} \right]$$
$$\Rightarrow {\text{y = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{{{\left( {\sqrt {1 + \sin {\text{x}}} + \sqrt {1 - \sin {\text{x}}} } \right)}^2}}}{{1 + \sin {\text{x}} - {\text{1 + sin x}}}}} \right)$$ $$\textbf{(Rationalizing the denominator)}$$
$$\Rightarrow {\text{y = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{1 + \sin x + 1 - \sin x + 2\sqrt {1 - \sin x} .\sqrt {1 + \sin x} }}{{2\sin x}}} \right)$$
$$\Rightarrow {\text{y = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{2\left( {1 + \cos x} \right)}}{{2\sin x}}} \right)$$
$$\Rightarrow {\text{y = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}.\cos \dfrac{x}{2}}}} \right)$$
$$\Rightarrow {\text{y = ta}}{{\text{n}}^{ - 1}}\left( {\cot \dfrac{x}{2}} \right)$$
$$\Rightarrow y = \dfrac{\pi }{2} - {\cot ^{ - 1}}\cot \dfrac{x}{2}$$
$$\Rightarrow {\text{y = }}\dfrac{\pi }{2} - \dfrac{x}{2}$$
$${\textbf{Step -2: Differentiating the obtained Expression w.r.t x}}{\textbf{.}}$$
$$\therefore\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{2}$$
$${\textbf{ Hence, the value of }} {\mathbf{\dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{2}.}}$$

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