tan ^{-1} left( frac{cos x}{1-sin x} right),- | vedclass

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(D) આપેલ પદ: $	an ^{-1} \left( \frac{\cos x}{1-\sin x} ight)$ત્રિકોણમિતીય નિત્યસમ $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$,$\sin x = 2 \s

Vedclass · Accepted Answer

$\frac{\pi}{4} + \frac{x}{2}$

Vedclass · Answer

(D) આપેલ પદ: $	an ^{-1} \left( \frac{\cos x}{1-\sin x} ight)$ત્રિકોણમિતીય નિત્યસમ $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}$,$\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}$,અને $1 = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}$ નો ઉપયોગ કરતા:$= 	an ^{-1} \left[ \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} - 2 \sin \frac{x}{2} \cos \frac{x}{2}} ight]$$= 	an ^{-1} \left[ \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})}{(\cos \frac{x}{2} - \sin \frac{x}{2})^2} ight]$$= 	an ^{-1} \left( \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}} ight)$અંશ અને છેદને $\cos \frac{x}{2}$ વડે ભાગતા:$= 	an ^{-1} \left( \frac{1 + 	an \frac{x}{2}}{1 - 	an \frac{x}{2}} ight) = 	an ^{-1} \left[ 	an \left( \frac{\pi}{4} + \frac{x}{2} ight) ight]$$= \frac{\pi}{4} + \frac{x}{2}$

$\tan ^{-1} \left( \frac{\cos x}{1-\sin x} \right)$,$-\frac{3 \pi}{2} < x < \frac{\pi}{2}$ ને સાદા સ્વરૂપમાં દર્શાવો.

Similar Questions

જો $\cos^{-1} p + \cos^{-1} q + \cos^{-1} r = \pi$ હોય,તો $p^2 + q^2 + r^2 + 2pqr = $

$\tan \left(2 \tan ^{-1} \frac{1}{5} + \sec ^{-1} \frac{\sqrt{5}}{2} + 2 \tan ^{-1} \frac{1}{8}\right)$ ની કિંમત શોધો.

જો $\frac{1}{2} \leq x \leq 1$ હોય,તો $\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{1}{2} \sqrt{3-3 x^2}\right)$ ની કિંમત શોધો.

સાબિત કરો કે $\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}$,જ્યાં $x \in\left(0, \frac{\pi}{4}\right)$.

જો $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$ હોય,તો $x$ ની કિંમત શોધો.

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