Question

Which of the following is/are true ?

• A. $\frac{dy}{dx}$ for $y={\sin }^{-1}(\cos x)$, where $x\in (0,\pi )$, is $-1$
• B. $\frac{dy}{dx}$ for $y={\sin }^{-1}(\cos x)$, where $x\in (\pi ,2\pi )$, is 1
• C. $\frac{dy}{dx}$ for $y={\cos }^{-1}(\sin x)$, where $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$, is $-1$
• D. $\frac{dy}{dx}$ for $y={\cos }^{-1}(\sin x)$, where $x\in \left(\frac{\pi }{2},\frac{3\pi }{2}\right)$, is $-1$

Answer 1

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

We have ${\sin }^{-1}(\cos x)=\frac{\pi }{2}-{\cos }^{-1}(\cos x)$
$=\{\begin{array}{ll}\frac{\pi }{2}-x,& \text{ if }0<x\le \pi \\ \frac{\pi }{2}-(2\pi -x),& \text{ if }\pi <x<2\pi \end{array}$
$=\{\begin{array}{ll}\frac{\pi }{2}-x,& \text{ if }0<x\le \pi \\ <br/><br/>x-\frac{3\pi }{2},& \text{ if }\pi <x<2\pi \end{array}$
$\therefore \frac{d}{dx}\left\{{\sin }^{-1}(\cos x)\right\}=\{\begin{array}{ll}-1,& \text{ if }0<x<\pi \\ 1,& \text{ if }\pi <x<2\pi \end{array}$
We have ${\cos }^{-1}(\sin x)=\frac{\pi }{2}-{\sin }^{-1}(\sin x)$
$=\{\begin{array}{ll}\frac{\pi }{2}-x,& \text{ if }-\frac{\pi }{2}<x\le \frac{\pi }{2}\\ \frac{\pi }{2}-(\pi -x),& \text{ if }\frac{\pi }{2}<x<\frac{3\pi }{2}\end{array}$
$=\{\begin{array}{ll}\frac{\pi }{2}-x,& \text{ if }-\frac{\pi }{2}<x\le \frac{\pi }{2}\\ x-\frac{\pi }{2},& \text{ if }\frac{\pi }{2}<x<\frac{3\pi }{2}\end{array}$
$\therefore \frac{d}{dx}\left({\cos }^{-1}(\sin x)\right)=\{\begin{array}{ll}-1,& \text{ if }-\frac{\pi }{2}<x<\frac{\pi }{2}\\ 1,& \text{ if }\frac{\pi }{2}<x<\frac{3\pi }{2}\end{array}$