Inverse Trigonometric Functions Question 1
Question: $ {{\cot }^{-1}}[ \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}} ]= $
[MNR 1986]
Options:
A) $ \pi -x $
B) $ 2\pi -x $
C) $ \frac{x}{2} $
D) $ \pi -\frac{x}{2} $
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Answer:
Correct Answer: D
Solution:
$ {{\cot }^{-1}}[ \frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}} ] $
$ ={{\cot }^{-1}}[ \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})} ] $
= $ {{\cot }^{-1}}[ \frac{(1-\sin x)+(1+\sin x)+2\sqrt{1-{{\sin }^{2}}x}}{(1-\sin x)-(1+\sin x)} ] $
= $ {{\cot }^{-1}}[ \frac{2(1+\cos x)}{-2\sin x} ]={{\cot }^{-1}}[ -\frac{2{{\cos }^{2}}(x/2)}{2\sin (x/2)\cos (x/2)} ] $ = $ {{\cot }^{-1}}( -\cot \frac{x}{2} )={{\cot }^{-1}}[ \cot ( \pi -\frac{x}{2} ) ]=\pi -\frac{x}{2} $ .
Trick: Put $ x=\frac{\pi }{4} $ ,
so that the expression becomes $ {{\cot }^{-1}}[ \frac{\sqrt{\sqrt{2}-1}+\sqrt{\sqrt{2}+1}}{\sqrt{\sqrt{2}-1}-\sqrt{\sqrt{2}+1}} ] $
= $ {{\cot }^{-1}}[ \frac{\sqrt{2}-1+\sqrt{2}+1+2\sqrt{2-1}}{\sqrt{2}-1-\sqrt{2}-1} ] $
= $ {{\cot }^{-1}}[ \frac{2\sqrt{2}+2}{-2} ]={{\cot }^{-1}}(-1-\sqrt{2})=157.5{}^\circ $ .
BETA
