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