frac{{sin theta }}{{1 - cot theta }} + frac{{ | vedclass

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

Question

(D) Given expression: $\frac{{\sin 	heta }}{{1 - \cot 	heta }} + \frac{{\cos 	heta }}{{1 - 	an 	heta }}$Substitute $\cot 	heta = \frac{{\cos 	h

Vedclass · Accepted Answer

$\cos 	heta + \sin 	heta $

Vedclass · Answer

(D) Given expression: $\frac{{\sin 	heta }}{{1 - \cot 	heta }} + \frac{{\cos 	heta }}{{1 - 	an 	heta }}$Substitute $\cot 	heta = \frac{{\cos 	heta }}{{\sin 	heta }}$ and $	an 	heta = \frac{{\sin 	heta }}{{\cos 	heta }}$:$= \frac{{\sin 	heta }}{{1 - \frac{{\cos 	heta }}{{\sin 	heta }}}} + \frac{{\cos 	heta }}{{1 - \frac{{\sin 	heta }}{{\cos 	heta }}}}$$= \frac{{\sin 	heta }}{{\frac{{\sin 	heta - \cos 	heta }}{{\sin 	heta }}}} + \frac{{\cos 	heta }}{{\frac{{\cos 	heta - \sin 	heta }}{{\cos 	heta }}}}$$= \frac{{{{\sin }^2}	heta }}{{\sin 	heta - \cos 	heta }} + \frac{{{{\cos }^2}	heta }}{{\cos 	heta - \sin 	heta }}$$= \frac{{{{\sin }^2}	heta }}{{\sin 	heta - \cos 	heta }} - \frac{{{{\cos }^2}	heta }}{{\sin 	heta - \cos 	heta }}$$= \frac{{{{\sin }^2}	heta - {{\cos }^2}	heta }}{{\sin 	heta - \cos 	heta }}$Using the identity $a^2 - b^2 = (a - b)(a + b)$:$= \frac{{(\sin 	heta - \cos 	heta )(\sin 	heta + \cos 	heta )}}{{\sin 	heta - \cos 	heta }}$$= \sin 	heta + \cos 	heta $.

$\frac{{\sin \theta }}{{1 - \cot \theta }} + \frac{{\cos \theta }}{{1 - \tan \theta }} = $

Similar Questions

If $\tan \theta - \cot \theta = a$ and $\sin \theta + \cos \theta = b,$ then ${({b^2} - 1)^2}({a^2} + 4)$ is equal to

For $A = 133^\circ$,$2\cos \frac{A}{2}$ is equal to

Given that $\pi < \alpha < \frac{3\pi}{2},$ then the expression $\sqrt{4\sin^4 \alpha + \sin^2 2\alpha} + 4\cos^2 \left( \frac{\pi}{4} - \frac{\alpha}{2} \right)$ is equal to

If $\sin 2\theta + \sin 2\phi = 1/2$ and $\cos 2\theta + \cos 2\phi = 3/2$,then $\cos^2(\theta - \phi) = $

$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot 8\alpha = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module