(sec x – 1) (sec x + 1)


(sec x – 1) (sec x + 1)


Given $(\sec x-1)(\sec x+1)$

Let $y=(\operatorname{secx}-1)(\operatorname{secx}+1)$

The above equation can be written as

$\Rightarrow y=(\sec x-1)(\sec x+1)=\sec ^{2} x-1=\tan ^{2} x$

$\Rightarrow y=\tan ^{2} x$

Now applying the chain rule we get

$\Rightarrow \frac{d y}{d x}=\frac{d}{d(\tan x)}\left(\tan ^{2} x\right) \cdot \frac{d}{d x}(\tan x)$

$\Rightarrow \frac{d y}{d x}=2 \tan x \sec ^{2} x$

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