A) \[\frac{x}{2}\]
B) \[\pi -\frac{x}{2}\]
C) \[\pi -x\]
D) \[2\pi -x\]
Correct Answer: B
Solution :
Given, \[{{\cot }^{-1}}\left[ \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}} \right]\] \[={{\cot }^{-1}}\left[ \frac{\begin{align} & \sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}-2\sin \frac{x}{2}.\cos \frac{x}{2}} \\ & +\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}} \\ \end{align}}{\begin{align} & \sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}-2\sin \frac{x}{2}.\cos \frac{x}{2}} \\ & -\sqrt{{{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}} \\ \end{align}} \right]\] \[={{\cot }^{-1}}\left[ \frac{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)+\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)-\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)} \right]\] \[={{\cot }^{-1}}\left[ \frac{2\cos x/2}{-2\sin x/2} \right]\] \[={{\cot }^{-1}}\left[ -{{\cot }^{-1}}\frac{x}{2} \right]\] \[={{\cot }^{-1}}\left[ \cot \left( \pi -\frac{x}{2} \right) \right]=\pi -\frac{x}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
