Inverse Trigonometric Functions Question 51
Question: If $ x\in [\pi /2,\pi ] $ then $ {{\cot }^{-1}}( \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} )= $
Options:
A) $ \frac{x-\pi }{2} $
B) $ \frac{\pi -x}{2} $
C) $ \frac{3\pi -x}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
$ \sqrt{1+\sin x}=\sin \frac{x}{2}+\cos \frac{x}{2} $
$ \sqrt{1-\sin x}=\sin \frac{x}{2}-\cos \frac{x}{2} $
$ [ for\frac{\pi }{4}\le \frac{x}{2}\le \frac{\pi }{2}\sin \frac{\pi }{2}\ge \cos \frac{x}{2} ] $
$ \therefore $ the expression is $ {{\cot }^{-1}}( \frac{\sin \frac{x}{2}+\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}} ) $
$ ={{\cot }^{-1}}( \tan \frac{x}{2} )={{\cot }^{-1}}\cot ( \frac{\pi }{2}-\frac{x}{2} )=\frac{\pi -x}{2} $
BETA
