Question

$co{t}^{-1}\frac{[\sqrt{1-\sin x}+\sqrt{1+\sin x}]}{[\sqrt{1-\sin x}-\sqrt{1+\sin x}]}=?$

• A. $\mathrm{\pi }-\mathrm{x}$
• B. $2\mathrm{\pi }-\mathrm{x}$
• C. $\frac{x}{2}$
• D. $\mathrm{\pi }-\frac{\mathrm{x}}{2}$

Answer 1

The correct option is D

$\mathrm{\pi }-\frac{\mathrm{x}}{2}$

Explanation for the correct option:

Given, $co{t}^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$

By rationalizing it, we get

$=co{t}^{-1}\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]\times \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}+\sqrt{1+\sin x}}$

$=co{t}^{-1}\left[\frac{{\left(\sqrt{1-\sin x}+\sqrt{1+\sin x}\right)}^2}{\left(1-\sin x\right)-\left(1+\sin x\right)}\right]$

$=co{t}^{-1}\left(\frac{1-\sin x+1+\sin x+2\sqrt{1-{\sin }^{2}x}}{1-\sin x-1-\sin x}\right)$

$=co{t}^{-1}\left(\frac{2+2\sqrt{{\cos }^2}x}{-2\sin x}\right)$

$=co{t}^{-1}\left(\frac{1+\cos x}{-\sin x}\right)$

$=co{t}^{-1}\left(\frac{1+2{\cos }^2{\displaystyle \frac{x}{2}}-1}{-2\sin {\displaystyle \frac{x}{2}}\cos {\displaystyle \frac{x}{2}}}\right)$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\frac{\mathbf{x}}{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\frac{\mathbf{x}}{\mathbf{2}}\mathit{c}\mathit{o}\mathit{s}\frac{\mathbf{x}}{\mathbf{2}}\right]$

$=co{t}^{-1}\left(\frac{\cos {\displaystyle \frac{x}{2}}}{-\sin {\displaystyle \frac{x}{2}}}\right)$

$=co{t}^{-1}\left(-cot\frac{x}{2}\right)$

$=co{t}^{-1}\left[cot\left(\pi -\frac{x}{2}\right)\right]$ ; $\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{t}\left(\pi -\frac{x}{2}\right)\mathbf{=}\mathbf{-}\mathit{c}\mathit{o}\mathit{t}\frac{\mathbf{x}}{\mathbf{2}}\right]$

$\mathbf{=}\mathit{\pi }\mathbf{-}\frac{\mathbf{x}}{\mathbf{2}}$

Hence, Option ‘D’ is Correct.