A) \[\pi -x\]
B) \[2\pi -x\]
C) \[\frac{x}{2}\]
D) \[\pi -\frac{x}{2}\]
Correct Answer: D
Solution :
\[{{\cot }^{-1}}\left[ \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}} \right]\] \[={{\cot }^{-1}}\left[ \frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}-\sqrt{1+\sin x})}.\frac{(\sqrt{1-\sin x}+\sqrt{1+\sin x})}{(\sqrt{1-\sin x}+\sqrt{1+\sin x})} \right]\] =\[{{\cot }^{-1}}\left[ \frac{(1-\sin x)+(1+\sin x)+2\sqrt{1-{{\sin }^{2}}x}}{(1-\sin x)-(1+\sin x)} \right]\] =\[{{\cot }^{-1}}\left[ \frac{2(1+\cos x)}{-2\sin x} \right]={{\cot }^{-1}}\left[ -\frac{2{{\cos }^{2}}(x/2)}{2\sin (x/2)\cos (x/2)} \right]\] =\[{{\cot }^{-1}}\left( -\cot \frac{x}{2} \right)={{\cot }^{-1}}\left[ \cot \left( \pi -\frac{x}{2} \right) \right]=\pi -\frac{x}{2}\]. Trick: Put\[x=\frac{\pi }{4}\], so that the expression becomes \[{{\cot }^{-1}}\left[ \frac{\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}} \right]\] = \[{{\cot }^{-1}}\left[ \frac{\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}}{\sqrt{2}-1-\sqrt{2}-1} \right]\] = \[{{\cot }^{-1}}\left[ \frac{2\sqrt{2}+2}{-2} \right]={{\cot }^{-1}}(-1-\sqrt{2})=157.5{}^\circ \].
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