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

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(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

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