જો y = cos^2 [cot^{-1} (sqrt{frac{1-x}{1+x}}) | vedclass

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(D) ધારો કે $x = \cos 	heta$, તો $	heta = \cos^{-1} x$ થાય.$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{2\sin^

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(D) ધારો કે $x = \cos 	heta$, તો $	heta = \cos^{-1} x$ થાય.$\sqrt{\frac{1-x}{1+x}} = \sqrt{\frac{1-\cos 	heta}{1+\cos 	heta}} = \sqrt{\frac{2\sin^2(	heta/2)}{2\cos^2(	heta/2)}} = 	an(	heta/2)$ થાય.હવે, $\cot^{-1}(	an(	heta/2)) = \cot^{-1}(\cot(\frac{\pi}{2} - \frac{	heta}{2})) = \frac{\pi}{2} - \frac{	heta}{2}$ થાય.આને $y$ ના પદમાં મૂકતા:$y = \cos^2(\frac{\pi}{2} - \frac{	heta}{2}) = \sin^2(	heta/2)$ થાય.નિત્યસમ $\sin^2(	heta/2) = \frac{1-\cos 	heta}{2}$ નો ઉપયોગ કરતા:$y = \frac{1-x}{2} = \frac{1}{2} - \frac{x}{2}$ થાય.તેથી, $\frac{dy}{dx} = \frac{d}{dx}(\frac{1}{2} - \frac{x}{2}) = -\frac{1}{2}$ થાય.

જો $y = \cos^2 [\cot^{-1} (\sqrt{\frac{1-x}{1+x}})]$ હોય, તો $\frac{dy}{dx} = \dots$

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