Prove the following


$\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2}


Let $y=\tan ^{-1}(\sec x+\tan x)$

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=\frac{d}{d x}\left[\tan ^{-1}(\sec x+\tan x)\right]$

$=\frac{1}{1+(\sec x+\tan x)^{2}} \cdot \frac{d}{d x}(\sec x+\tan x)$

$=\frac{1}{1+\sec ^{2} x+\tan ^{2} x+2 \sec x \tan x} \cdot\left(\sec x \tan x+\sec ^{2} x\right)$

$=\frac{1}{\left(1+\tan ^{2} x\right)+\sec ^{2} x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x)$

$=\frac{1}{\sec ^{2} x+\sec ^{2} x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x)$

$=\frac{1}{2 \sec ^{2} x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x)$

$=\frac{1}{2 \sec x(\sec x+\tan x)} \cdot \sec x(\tan x+\sec x)=\frac{1}{2}$

Thus, $\quad \frac{d y}{d x}=\frac{1}{2}$

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