If sin x=

Solution:

Given $\sin x=\frac{2 t}{1+t^{2}}$ and $x$ lies in 2 nd quadrant

Since sin2x + cos2x = 1

cos2x = 1 – sin2x

Since x lies in II Quadrant

⇒ cos x < 0

$\therefore \cos x=-\sqrt{1-\sin ^{2} \theta}$

$=-\sqrt{1-\left(\frac{2 t}{1+t^{2}}\right)^{2}}$

$=-\sqrt{1-\frac{4 t^{2}}{\left(1+t^{2}\right)^{2}}}$

 

$=-\sqrt{\frac{\left(1+t^{2}\right)^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}}$

$=-\sqrt{\frac{1+t^{4}+2 t^{2}-4 t^{2}}{\left(1+t^{2}\right)^{2}}}$

 

$=-\sqrt{\frac{1+t^{4}-2 t^{2}}{\left(1+t^{2}\right)^{2}}}$

$=-\sqrt{\frac{\left(1-t^{2}\right)^{2}}{\left(1+t^{2}\right)^{2}}}$

Hence, $\cos x=-\frac{\left(1-t^{2}\right)}{1+t^{2}}$