Inverse Trigonometric Functions Question 123
Question: The value of $ {{\cos }^{-1}}x+{{\cos }^{-1}}( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} );\frac{1}{2}\le x\le 1 $ is
Options:
A) $ -\frac{\pi }{3} $
B) $ \frac{\pi }{3} $
C) $ \frac{3}{\pi } $
D) $ -\frac{3}{\pi } $
Show Answer
Answer:
Correct Answer: B
Let $ {{\cos }^{-1}}x=y $
$ \Rightarrow x=\cos y, $ so that $ \frac{1}{2}\le x\le 1 $ or $ 0\le y\le \frac{\pi }{3} $ and $ \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}}=\frac{1}{2}\cos y+\frac{\sqrt{3}}{2}\sin y $
$ =\cos \frac{\pi }{3}\cos y+\sin \frac{\pi }{3}\sin y=\cos ( \frac{\pi }{3}-y ) $
$ \Rightarrow {{\cos }^{-1}}( \frac{x}{2}+\frac{1}{2}\sqrt{3-3x^{2}} )=\frac{\pi }{3}-y $
$ \therefore $ the given expression is equal to $ y+\frac{\pi }{3}-y,i.e,\frac{\pi }{3} $
BETA
