Question

Consider the following statements
I : The solution set of the equation
$si{n}^{-1}(1-x)-2{\sin }^{-1}x=\frac{\pi }{2}$ is $\left\{0\right\}$
II :
$\tan \left\{Co{t}^{-1}\right(\frac{1}{5})+\frac{\pi }{4}\}=-\frac{3}{2}$
III :
${\tan }^{-1}\left[\right(\tan (-6)\left)\right]=-6$ Then which of the following is true

• A. only III
• B. both I and II
• C. both I and III
• D. I, II, III

Answer 1

The correct option is B both I and II

I. $si{n}^{-1}(1-x)-2si{n}^{-1}x=\frac{\pi }{2}$
domain $xϵ[-1,1]$ intersection $xϵ[0,2]=xϵ[0,1]$
$-1\le 1-x\le +1$
$-2\le -x\le 0$
$0\le x\le 2$
$si{n}^{-1}(1-x)=\frac{\pi }{2}+2si{n}^{-1}x$
take sin on both sides
$1-x=cos\left(2si{n}^{-1}x\right)$
$1-x=1-2\left(x^2\right)[cos2\theta =1-2si{n}^{2}x,sinsi{n}^{-1}x=x]$
$1-x=1-2x^2$
$\Rightarrow x=0$
II. $tan\left\{co{t}^{-1}\right(\frac{1}{5})+\frac{\pi }{4}\}=-\frac{3}{2}\left[tan\right(A+B)=\frac{tanA+tanB}{1-tanAtanB}]$
$\frac{1+5}{1-5}=\frac{6}{-4}=\frac{-3}{2}$ True
III. $ta{n}^{-1}\left[\right(tan(-6)\left)\right]=-ta{n}^{-1}tan6\frac{3\pi }{2}<6<2\pi$
$=-(-6)=6\Rightarrow tan6<0$ False
Both I and II are true.