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Q.
If the range of the function $f(x)=\sqrt{\tan ^{-1} x+1}+\sqrt{1-\tan ^{-1} x}$ is $[a, b]$ then find the value of $\left(a^2+b^2\right)$.
Inverse Trigonometric Functions
Solution:
$\text { For } y \text { to the real }-1 \leq \tan ^{-1} x \leq 1 $
$\Rightarrow -\tan 1 \leq x \leq \tan 1 $
$y=\sqrt{t+1}+\sqrt{1-t} \Rightarrow y^2=2+2 \sqrt{1-t^2} $
$\Rightarrow y^2 \in[2,4] \Rightarrow y \in[\sqrt{2}, 2]$
$\Rightarrow\left(a^2+b^2\right)=6$