frac{d}{dx} left( tan^{-1} left( frac{cos x}{ | vedclass

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

Question

(B) माना $y = 	an^{-1} \left( \frac{\cos x}{1 + \sin x} ight)$ है।त्रिकोणमितीय सर्वसमिकाओं $\cos x = \sin(\frac{\pi}{2} - x)$ और $1 + \sin x = 1 +

Vedclass · Accepted Answer

$-1/2$

Vedclass · Answer

(B) माना $y = 	an^{-1} \left( \frac{\cos x}{1 + \sin x} ight)$ है।त्रिकोणमितीय सर्वसमिकाओं $\cos x = \sin(\frac{\pi}{2} - x)$ और $1 + \sin x = 1 + \cos(\frac{\pi}{2} - x) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2})$ का उपयोग करने पर:$\frac{\cos x}{1 + \sin x} = \frac{\sin(\frac{\pi}{2} - x)}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = \frac{2 \sin(\frac{\pi}{4} - \frac{x}{2}) \cos(\frac{\pi}{4} - \frac{x}{2})}{2 \cos^2(\frac{\pi}{4} - \frac{x}{2})} = 	an(\frac{\pi}{4} - \frac{x}{2})$.अतः,$y = 	an^{-1} \left( 	an(\frac{\pi}{4} - \frac{x}{2}) ight) = \frac{\pi}{4} - \frac{x}{2}$.$x$ के सापेक्ष अवकलन करने पर:$\frac{dy}{dx} = \frac{d}{dx} \left( \frac{\pi}{4} - \frac{x}{2} ight) = 0 - \frac{1}{2} = -\frac{1}{2}$.

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{\cos x}{1 + \sin x} \right) \right) =$

Similar Questions

$\tan \left(\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ का मान है

यदि ${\tan ^{ - 1}}(x - 1) + {\tan ^{ - 1}}x + {\tan ^{ - 1}}(x + 1) = {\tan ^{ - 1}}3x$ है,तो $x =$

$\sin ^{-1}(\cos(\sin ^{-1} x)) + \cos ^{-1}(\sin(\cos^{-1} x)) = \text{ . . . . . . }$

मान ज्ञात कीजिए: $\tan ^2(\sec ^{-1} 3) + \operatorname{cosec}^2(\cot ^{-1} 2) + \cos ^2(\cos ^{-1} \frac{2}{3} + \sin ^{-1} \frac{2}{3}) = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module