Question

find derivative of the function:cot inverse(1+cosx/sinx)

Answer 1

here's the differentiation

Answer 2

$\frac{d}{dx} cot^{-1} \Big( \frac{1+cos x}{sin x } \Big)$

$= \frac{d}{dx} cot^{-1} \Big( \frac{2 cos ^{2} \frac{x}{2}}{2 sin \frac{x}{2} cos \frac{x}{2} }\Big)$

/* ___________________

$\pink { i ) 1 + Cos 2x = 2 cos^{2} x }$

$\blue { ii ) sin 2x = 2 sin x cos x }$

_________________ */

$= \frac{d}{dx} cot^{-1} \Big( \frac{cos \frac{x}{2}}{ sin \frac{x}{2} }\Big)$

$= \frac{d}{dx} cot^{-1} \Big( cot \frac{x}{2}\Big)$

$= \frac{d}{dx} \big(\frac{x}{2}\big) \\= \frac{1}{2}$

Therefore.,

$\red { Value \:of \:\frac{d}{dx} cot^{-1} \Big( \frac{1+cos x}{sin x }\Big)}$

$\green {= \frac{1}{2}}$

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