$(\tan x+\sec x)(\cot x+\operatorname{cosec} x)$



To find: Differentiation of $(\tan x+\sec x)(\cot x+\operatorname{cosec} x)$

Formula used: (i) (uv) $^{\prime}=u^{\prime} v+u v^{\prime}$ (Leibnitz or product rule)


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


$\frac{d \sec x}{d x}=\sec x \tan x$


$\frac{d \cot x}{d x}=-\operatorname{cosec}^{2} x$


$\frac{d \operatorname{cosec} x}{d x}=-\operatorname{cosec} x \cot x$

Let us take $u=(\tan x+\sec x)$ and $v=(\cot x+\operatorname{cosec} x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d(\tan x+\sec x)}{d x}=\sec ^{2} x+\sec x \tan x=\sec x(\sec x+\tan x)$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\cot x+\operatorname{cosec} x)}{d x}$

$=-\operatorname{cosec}^{2} x+(-\operatorname{cosec} x \cot x)=-\operatorname{cosec} x(\operatorname{cosec} x+\cot x)$

Putting the above obtained values in the formula:-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$=[\sec x(\sec x+\tan x)] \times[(\cot x+\operatorname{cosec} x)]+[(\tan x+\sec x)] \times[-\operatorname{cosec} x(\operatorname{cosec} x+\cot x)]$

$=(\sec x+\tan x)[\sec x(\cot x+\operatorname{cosec} x)-\operatorname{cosec} x(\operatorname{cosec} x+\cot x)]$

$=(\sec x+\tan x)(\sec x-\operatorname{cosec} x)(\cot x+\operatorname{cosec} x)$

Ans) $(\sec x+\tan x)(\sec x-\operatorname{cosec} x)(\cot x+\operatorname{cosec} x)$


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