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Q.
The domain set of definition of the function $f(x)=\sqrt{\cos (\sin x)}+\sin ^{-1}\left(\frac{1+x^2}{2 x}\right)$ is
Inverse Trigonometric Functions
Solution:
$ \cos (\sin x) \geq 0 \forall x, x \neq 0$
Hence, $\cos (\sin x)>0 \forall x$
$\frac{-\pi}{2} \leq \sin ^{-1}\left(\frac{1+x^2}{2 x}\right) \leq \frac{\pi}{2}$
i.e. $-1 \leq \frac{1+x^2}{2 x} \leq 1$. If $x<0$, then $\frac{1+x^2}{2 x}<0$
i.e. $-1 \leq \frac{1+x^2}{2 x} \Rightarrow-2 x \geq\left(1+x^2\right) \Rightarrow\left(1+x^2\right) \leq 0 \Rightarrow x=-1$
The other side $(x>0), 1+x^2 \leq 2 x \Rightarrow(x-1)^2 \leq 0 \Rightarrow x=1$.
The function is defined only for the two values $x=1$ and $x=-1$.