Inverse trig Range

$sin^{-1}x + cos^{-1} = \frac{\pi}{2}$

Domain is $\{-1 , 1\}$

Thus, $\boxed{\frac{-\pi}{4} \leq tan^{-1}x \leq \frac{\pi}{4}}$

Minimum - $\frac{pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$

Maximum - $\frac{\pi}{2} + \frac{\pi}{4} = \frac{3\pi}{4}$

Range - $\boxed{[\frac{\pi}{4} , \frac{3\pi}{4}]}$

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We cannot directly consider the function as $f(x) = \frac{\pi}{2} + tan^{-1}x$ .

[Note - we are changing the function here , which is a wrong step] . By the definition of function, the function given in the question will have restricted domain.

f(x) is a strictly increasing function( because f ' (x)= $\frac {1}{1 + x^2}$ > 0 for every x in (-1,1) and the domain of this function is [-1 ,1] where it is a continous function) $\Rightarrow$ f(x) will reach its minimun and its maximum in x=-1 and x=1 respectively $\Rightarrow$ f(-1) = $\pi$ /4 and f(1) = 3 $\pi$ /4 are the minimum and maximum respectively of its range $\Rightarrow$ the range of this function is the aforementioned.