Question

Differentiate - Tan^-1 [(x sin alpha) /(1-x cos alpha) ]

Answer 1

Step-by-step explanation:

in given picture there is correct answer of question given above

Answer 2

Answer:

$\frac{dy}{dx}=\frac{sin\alpha }{1-2xcos\alpha +x^2}$

Given:

Differentiate - Tan^-1 [(x sin alpha) /(1-x cos alpha) ]

To find :

Differentiate - Tan^-1 [(x sin alpha) /(1-x cos alpha) ]

Solution :

$y= tan^-1 (\frac{xsin\alpha }{1-xcos\alpha } )$

Differentiate with respect to x

$\frac{dy}{dx} =\frac{1}{1+(\frac{xsin\alpha }{1-xcos\alpha } )^2} *\frac{d}{dx}(\frac{xsin\alpha }{1-xcos\alpha } )$

$\frac{dy}{dx}=\frac{1}{1+\frac{x^2sin\alpha }{(1-xcos\alpha )^2} } *\frac{(1-xcos\alpha)\frac{d}{dx}(xsin\alpha )-[(xsin\alpha )\frac{d}{dx} (1-xcos\alpha )] }{1-xcos\alpha }$

$\frac{dy}{dx}=\frac{(1-xcos\alpha )^2}{(1-xcos\alpha )^2+x^2sin^2\alpha }*\frac{(1-xcos\alpha )sin\alpha -[(xsin\alpha )(0-cos\alpha )}{(1-xcos\alpha )^2}$

$\frac{dy}{dx}=\frac{sin\alpha -xcos\alpha .sin\alpha +xcos\alpha .sin\alpha }{1+x^2cos^2\alpha -2xcos\alpha +x^2sin^2\alpha }$

$\frac{dy}{dx}=\frac{sin\alpha }{1-2xcos\alpha +x^2(cos^2\alpha +sin^2\alpha )}$

$\frac{dy}{dx}=\frac{sin\alpha }{1-2xcos\alpha +x^2}$

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