Inverse Trigonometric Functions Question 61
Question: If $ {{\sin }^{-1}}(x-1)+co{s^{-1}}(x-3)+ta{n^{-1}}( \frac{x}{2-x^{2}} ) $
$ ={{\cos }^{-1}}k+\pi , $ Then the value of k is
Options:
A) 1
B) $ -\frac{1}{\sqrt{2}} $
C) $ \frac{1}{\sqrt{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
$ {{\sin }^{-1}}(x-1)\Rightarrow -1\le x-1\le 1\Rightarrow 0\le x\le 2 $
$ {{\cos }^{-1}}(x-3)\Rightarrow -1\le x-3\le 1\Rightarrow 2\le x\le 4 $
$ \therefore x=2 $ So, $ {{\sin }^{-1}}(2-1)+co{s^{-1}}(2-3)+ta{n^{-1}}\frac{2}{2-4} $
$ ={{\cos }^{-1}}k+\pi $
Or $ {{\sin }^{-1}}1+{{\cos }^{-1}}(-1)+ta{n^{-1}}(-1)=co{s^{-1}}k+\pi $
$ \Rightarrow \frac{\pi }{2}+\pi -\frac{\pi }{4}={{\cos }^{-1}}k+\pi \Rightarrow {{\cos }^{-1}}k=\frac{\pi }{4} $ or $ k=\frac{1}{\sqrt{2}} $
BETA
