Question

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

• (A) $$\frac{1-t^2}{1+t^2}$$
• (B) $$\frac{t^2-1}{1+t^2}$$
• (C) $$\frac{-|1-t^2|}{1+t^2}$$
• (D) $$\frac{1+t^2}{|1-t^2|}$$

Answer 1

The Correct Option is C

Explanation:
θ in 2nd quad Cosθ < 0$$\begin{array}{l}|\cos \theta |=\left|\frac{1-t^2}{1+t^2}\right|=\frac{|1-t^2|}{1+t^2}\\ \cos \theta =-\frac{|1-t^2|}{1+t^2}\end{array}$$