Inverse Trigonometric Functions Question 220
Question: If $ a{{\sin }^{-1}}x-b{{\cos }^{-1}}x=c, $ then $ a{{\sin }^{-1}}x+b{{\cos }^{-1}}x $ is equal to
Options:
A) 0
B) $ \frac{\pi ab+c(b-c)}{a+b} $
C) $ \frac{\pi }{2} $
D) $ \frac{\pi ab+c(a-b)}{a+b} $
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Answer:
Correct Answer: D
$ a{{\sin }^{-1}}x-b{{\cos }^{-1}}x=c $
We have b $ {{\sin }^{-1}}x+b{{\cos }^{-1}}x=\frac{b\pi }{2} $ Adding $ (a+b) $
$ {{\sin }^{-1}}x=\frac{b\pi }{2}+c $
Or $ {{\sin }^{-1}}x=\frac{( \frac{b\pi }{2} )+c}{a+b}=\frac{b\pi +2c}{2(a+b)} $
$ \therefore {{\cos }^{-1}}x=\frac{\pi }{2}-\frac{b\pi +2c}{2(a+b)}=\frac{\pi a-2c}{2(a+b)} $
$ \Rightarrow a{{\sin }^{-1}}x+b{{\cos }^{1}}x=\frac{\pi ab+c(a-b)}{a+b} $
BETA
