tan ^{-1}left[frac{sin x}{1+cos x}right] નું | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) ધારો કે $u = 	an ^{-1}\left(\frac{\sin x}{1+\cos x}ight)$ અને $v = 	an ^{-1}\left(\frac{\cos x}{1+\sin x}ight)$.$u$ માટે:$u = 	an ^{-1}\lef

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) ધારો કે $u = 	an ^{-1}\left(\frac{\sin x}{1+\cos x}ight)$ અને $v = 	an ^{-1}\left(\frac{\cos x}{1+\sin x}ight)$.$u$ માટે:$u = 	an ^{-1}\left(\frac{2 \sin(x/2) \cos(x/2)}{2 \cos^2(x/2)}ight) = 	an ^{-1}(	an(x/2)) = x/2$.તેથી,$\frac{du}{dx} = \frac{1}{2}$.$v$ માટે:$v = 	an ^{-1}\left(\frac{\cos^2(x/2) - \sin^2(x/2)}{(\cos(x/2) + \sin(x/2))^2}ight) = 	an ^{-1}\left(\frac{(\cos(x/2) - \sin(x/2))(\cos(x/2) + \sin(x/2))}{(\cos(x/2) + \sin(x/2))^2}ight)$.$v = 	an ^{-1}\left(\frac{\cos(x/2) - \sin(x/2)}{\cos(x/2) + \sin(x/2)}ight) = 	an ^{-1}\left(\frac{1 - 	an(x/2)}{1 + 	an(x/2)}ight) = 	an ^{-1}\left(	an\left(\frac{\pi}{4} - \frac{x}{2}ight)ight) = \frac{\pi}{4} - \frac{x}{2}$.તેથી,$\frac{dv}{dx} = -\frac{1}{2}$.હવે,$\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/2}{-1/2} = -1$.

$\tan ^{-1}\left[\frac{\sin x}{1+\cos x}\right]$ નું $\tan ^{-1}\left[\frac{\cos x}{1+\sin x}\right]$ ની સાપેક્ષમાં વિકલન શું થાય?

Similar Questions

જો $y=\sin ^{-1}\left[x \sqrt{1-x^2}-\sqrt{x} \sqrt{1-x}\right]$ અને $0 < x < 1$ હોય,તો $\frac{d y}{d x}$ ની કિંમત શોધો.

જો $y=\tan ^{-1}\left(\frac{2-3 \sin x}{3-2 \sin x}\right)$ હોય,તો $\frac{d y}{d x}=$

જો $y=\tan ^{-1}\left(\frac{3+2 x}{2-3 x}\right)+\tan ^{-1}\left(\frac{3 x}{1+4 x^2}\right)$ હોય,તો $\frac{d y}{d x}$ ની કિંમત શોધો.

${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ નું ${\tan ^{ - 1}}x$ ની સાપેક્ષે વિકલન ગુણાંક શોધો.

જો $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right) + \sec^{-1} \left( \frac{1 + x^2}{1 - x^2} \right)$ હોય,તો $\frac{dy}{dx} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module