Question

The value of tan $\left[si{n}^{-1}cos\right(si{n}^{-1}x)+co{s}^{-1}sin(co{s}^{-1}x\left)\right],xϵ$ ($0$, 1) is equal to

• A. $1$
• B. $0$
• C. $2x$
• D. undefined

Answer 1

The correct option is D undefined

If $si{n}^{-1}x=\alpha ,co{s}^{-1}x=(\frac{\pi }{2}-\alpha )$
$\therefore si{n}^{-1}\left(cos\alpha \right)+co{s}^{-1}sin(\frac{\pi }{2}-\alpha )$

$\therefore tan(si{n}^{-1}\left(cos\left(si{n}^{-1}x\right)\right)+co{s}^{-1}sin\left(co{s}^{-1}x\right))$

$=tan(\frac{\pi }{2}-co{s}^{-1}cos\left(si{n}^{-1}x\right)+\frac{\pi }{2}-si{n}^{-1}sin\left(co{s}^{-1}x\right))$

$=tan(\pi -si{n}^{-1}x-co{s}^{-1}x)$

$=tan(\pi -(si{n}^{-1}x+co{s}^{-1}x))$

$=tan(\pi -\frac{\pi }{2})$

$=tan\left(\frac{\pi }{2}\right)=\infty$

Hence, it is undefined.