If y = cot^{-1} left( frac{1 + sin 5x}{cos 5x | vedclass

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

Question

(D) Given $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} ight)$.Using trigonometric identities $1 + \sin 	heta = \left( \cos \frac{	heta}{2} +

Vedclass · Accepted Answer

$-\frac{5}{2}$

Vedclass · Answer

(D) Given $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} ight)$.Using trigonometric identities $1 + \sin 	heta = \left( \cos \frac{	heta}{2} + \sin \frac{	heta}{2} ight)^2$ and $\cos 	heta = \cos^2 \frac{	heta}{2} - \sin^2 \frac{	heta}{2}$, we have:$y = \cot^{-1} \left( \frac{(\cos \frac{5x}{2} + \sin \frac{5x}{2})^2}{(\cos \frac{5x}{2} - \sin \frac{5x}{2})(\cos \frac{5x}{2} + \sin \frac{5x}{2})} ight)$$y = \cot^{-1} \left( \frac{\cos \frac{5x}{2} + \sin \frac{5x}{2}}{\cos \frac{5x}{2} - \sin \frac{5x}{2}} ight)$Divide numerator and denominator by $\cos \frac{5x}{2}$:$y = \cot^{-1} \left( \frac{1 + 	an \frac{5x}{2}}{1 - 	an \frac{5x}{2}} ight) = \cot^{-1} \left( 	an \left( \frac{\pi}{4} + \frac{5x}{2} ight) ight)$Since $\cot^{-1}(	an 	heta) = \cot^{-1}(\cot(\frac{\pi}{2} - 	heta))$, we get:$y = \frac{\pi}{2} - (\frac{\pi}{4} + \frac{5x}{2}) = \frac{\pi}{4} - \frac{5x}{2}$Therefore, $\frac{dy}{dx} = \frac{d}{dx} (\frac{\pi}{4} - \frac{5x}{2}) = -\frac{5}{2}$.

If $y = \cot^{-1} \left( \frac{1 + \sin 5x}{\cos 5x} \right)$, then the value of $\frac{dy}{dx}$ is

Similar Questions

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then

If $(\sin ^{-1} x)^{2}-(\cos ^{-1} x)^{2}=a ; 0 < x < 1, a \neq 0$,then the value of $2 x^{2}-1$ is :

If $\sin ^{ - 1}\frac{{2a}}{{1 + {a^2}}} - \cos ^{ - 1}\frac{{1 - {b^2}}}{{1 + {b^2}}} = \tan ^{ - 1}\frac{{2x}}{{1 - {x^2}}}$,then $x = $

If $\frac{\sin ^{-1} x}{a}=\frac{\cos ^{-1} x}{b}=\frac{\tan ^{-1} y}{c}$ and $0 < x < 1,$ then the value of $\cos \left(\frac{\pi c}{a + b}\right)$ is

The value of $\tan ^{-1}\left\{\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right\}+\frac{1}{2} \cos ^{-1} x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module