If sin theta = frac{2t}{1+t^2} and theta lies | vedclass

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

Question

(C) Given $\sin 	heta = \frac{2t}{1+t^2}$.We know that $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \left(\frac{2t}{1+t^2}ight)^2$.$\cos^2 	heta = 1 -

Vedclass · Accepted Answer

$\frac{-\left|1-t^2\right|}{1+t^2}$

Vedclass · Answer

(C) Given $\sin 	heta = \frac{2t}{1+t^2}$.We know that $\cos^2 	heta = 1 - \sin^2 	heta = 1 - \left(\frac{2t}{1+t^2}ight)^2$.$\cos^2 	heta = 1 - \frac{4t^2}{(1+t^2)^2} = \frac{(1+t^2)^2 - 4t^2}{(1+t^2)^2} = \frac{1 + 2t^2 + t^4 - 4t^2}{(1+t^2)^2} = \frac{1 - 2t^2 + t^4}{(1+t^2)^2} = \frac{(1-t^2)^2}{(1+t^2)^2}$.Taking the square root, $|\cos 	heta| = \frac{|1-t^2|}{1+t^2}$.Since $	heta$ lies in the second quadrant, $\cos 	heta$ must be negative.Therefore, $\cos 	heta = -\frac{|1-t^2|}{1+t^2}$.

If $\sin \theta = \frac{2t}{1+t^2}$ and $\theta$ lies in the second quadrant, then $\cos \theta$ is equal to

Similar Questions

$\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right) = $

${\left( \frac{\cos A + \cos B}{\sin A - \sin B} \right)^n} + {\left( \frac{\sin A + \sin B}{\cos A - \cos B} \right)^n}$ (where $n$ is an integer) $=$

If $A, B, C, D$ are the angles of a cyclic quadrilateral,then $\cos A + \cos B + \cos C + \cos D = $

$\sin ^4 \frac{\pi}{8}+\sin ^4 \frac{3 \pi}{8}+\sin ^4 \frac{5 \pi}{8}+\sin ^4 \frac{7 \pi}{8} = ?$

If $\cos \alpha + \cos \beta + \cos \gamma = 0$ and $\sin \alpha + \sin \beta + \sin \gamma = 0$,then $\cos 2\alpha + \cos 2\beta + \cos 2\gamma$ equals:

Vedclass Test Series

Exam Paper Generator

Online Exam Module