Question

If $y=se{c}^{-1}\left[\frac{x+1}{x-1}\right]+si{n}^{-1}\left[\frac{x-1}{x+1}\right]$ $z=cose{c}^{-1}\left[\frac{2x+3}{3x+2}\right]+co{s}^{-1}\left[\frac{3x+2}{2x+3}\right]$ then

• A. $y=\pi /2$
• B. $z=\pi /2$
• C. $y+z=\pi$
• D. $y+z=\pi /2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $z=\pi /2$
B $y=\pi /2$
C $y+z=\pi$

Given $y=se{c}^{-1}\left[\frac{x+1}{x-1}\right]+si{n}^{-1}\left[\frac{x-1}{x+1}\right],$ $z=cose{c}^{-1}\left[\frac{2x+3}{3x+2}\right]+co{s}^{-1}\left[\frac{3x+2}{2x+3}\right]$

$y=se{c}^{-1}\left[\frac{x+1}{x-1}\right]+si{n}^{-1}\left[\frac{x-1}{x+1}\right]$

$y={\cos }^{-1}\left(\frac{x-1}{x+1}\right)+{\sin }^{-1}\left(\frac{x-1}{x+1}\right)$

$=\frac{\pi }{2}$

$z=cose{c}^{-1}\left[\frac{2x+3}{3x+2}\right]+co{s}^{-1}\left[\frac{3x+2}{2x+3}\right]$

$z={\sin }^{-1}\left(\frac{3x+2}{2x+3}\right)+{\cos }^{-1}\left(\frac{3x+2}{2x+3}\right)$

$=\frac{\pi }{2}$

Hence

$y+z=\pi$