Question

\[ \text{If } \sin x=\frac{2t}{1+t^2} \]

\[ \text{and } x \text{ lies in the second quadrant,} \]

\[ \text{then } \cos x= \]

Solution

Using identity

\[ \sin^2x+\cos^2x=1 \]

\[ \cos^2x = 1-\left(\frac{2t}{1+t^2}\right)^2 \]

\[ = \frac{(1+t^2)^2-4t^2}{(1+t^2)^2} \]

\[ = \frac{1-2t^2+t^4}{(1+t^2)^2} \]

\[ = \frac{(1-t^2)^2}{(1+t^2)^2} \]

\[ \cos x = \pm\frac{1-t^2}{1+t^2} \]

Since \(x\) lies in second quadrant,

\[ \cos x<0 \]

Therefore,

\[ \cos x = -\frac{1-t^2}{1+t^2} = \frac{t^2-1}{1+t^2} \]

Answer

\[ \boxed{\frac{t^2-1}{1+t^2}} \]