Question:
Solve for x :
$\tan ^{-1} x=\sin ^{-1} \frac{1}{\sqrt{2}}$
Solution:
To find: value of x
Given: $\tan ^{-1} \mathrm{X}=\sin ^{-1} \frac{1}{\sqrt{2}}$
We know that $\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
Therefore, $\frac{\pi}{4}=\sin ^{-1} \frac{1}{\sqrt{2}}$
Substituting in the given equation,
$\tan ^{-1} x=\frac{\pi}{4}$
$x=\tan \frac{\pi}{4}$
$\Rightarrow x=1$
Therefore, x = 1 is the required value of x.
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