Inverse Trigonometric Functions Question 132
Question: The value of $ {{\sin }^{-1}}[cos{co{s^{-1}}(cosx)+si{n^{-1}}(sinx)}] $ where $ x\in ( \frac{\pi }{2},\pi ) $ is equal to
Options:
A) $ \frac{\pi }{2} $
B) $ -\pi $
C) $ \pi $
D) $ -\frac{\pi }{2} $
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Answer:
Correct Answer: D
$ {{\sin }^{-1}}[\cos {{{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)}] $
$ ={{\sin }^{-1}}[\cos (x+\pi -x)] $ as $ x\in (\pi /2,\pi ) $
$ ={{\sin }^{-1}}(cos\pi )=si{n^{-1}}(-1)=-\frac{\pi }{2} $
BETA
