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

Question

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

MARKS App · Accepted Answer

Hints: $\theta$ in 2nd quad $\cos \theta < 0$ $\begin{aligned} & |\cos \theta|=\left|\frac{1-t^2}{1+t^2}\right|=\frac{\left|1-t^2\right|}{1+t^2} \ & \cos \theta=-\frac{\left|1-t^2\right|}{1+t^2} \end{aligned}$

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