Question
Using derivative prove that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.
✓
Answer
Let $f(x)=\sin ^{-1} x+\cos ^{-1} x \quad \ldots .(I)$
We have to prove that $f(x)=\frac{\pi}{2}$
Differentiate (I) w.r.t.x
$
\begin{aligned}
\frac{d}{d x}[f(x)] & =\frac{d}{d x}\left[\sin ^{-1} x+\cos ^{-1} x\right] \\
f^{\prime}(x) & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0
\end{aligned}
$
$f^{\prime}(x)=0 \Rightarrow f(x)$ is a constant function.
Let $f(x)=c$. For any value of $x, f(x)$ must be $c$ only. So conveniently we can choose $x=0$,
$\therefore \quad$ from (I) we get,
$
f(0)=\sin ^{-1}(0)+\cos ^{-1}(0)=0+\frac{\pi}{2}=\frac{\pi}{2} \Rightarrow c=\frac{\pi}{2} \therefore f(x)=\frac{\pi}{2}
$
Hence, $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.
We have to prove that $f(x)=\frac{\pi}{2}$
Differentiate (I) w.r.t.x
$
\begin{aligned}
\frac{d}{d x}[f(x)] & =\frac{d}{d x}\left[\sin ^{-1} x+\cos ^{-1} x\right] \\
f^{\prime}(x) & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0
\end{aligned}
$
$f^{\prime}(x)=0 \Rightarrow f(x)$ is a constant function.
Let $f(x)=c$. For any value of $x, f(x)$ must be $c$ only. So conveniently we can choose $x=0$,
$\therefore \quad$ from (I) we get,
$
f(0)=\sin ^{-1}(0)+\cos ^{-1}(0)=0+\frac{\pi}{2}=\frac{\pi}{2} \Rightarrow c=\frac{\pi}{2} \therefore f(x)=\frac{\pi}{2}
$
Hence, $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.
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