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

Given $\sin \theta =\frac{2t}{1+t^2}$
Since, $\theta$ lies in the second quadrant
$\therefore$ $\cos \theta <0$
$\Rightarrow \cos \theta =-\sqrt{1-{\sin }^2\theta }$
$=-\sqrt{1-\frac{4t^2}{{(1+t^2)}^2}}$
$=-\sqrt{\frac{{(1-t^2)}^2}{{(1+t^2)}^2}}$
$=-\frac{|1-t^2|}{1+t^2}$