Question

If $\sin x=\frac{2t}{1+t^2}$ and x lies in the second quadrant, then cos x = ___________.

Answer 1

Given $\sin x=\frac{2t}{1+t^2}$ and x lies in 2nd quadrant
Since sin²x + cos²x = 1
cos²x = 1 – sin²x
Since x lies in II Quadrant
⇒ cos x < 0
$$\begin{gathered} \therefore \cos x=-\sqrt{1-{\sin }^2\theta } \\ =-\sqrt{1-{\left(\frac{2t}{1+t^2}\right)}^2} \\ =-\sqrt{1-\frac{4t^2}{{\left(1+t^2\right)}^2}} \\ =-\sqrt{\frac{{\left(1+t^2\right)}^2-4t^2}{{\left(1+t^2\right)}^2}} \\ =-\sqrt{\frac{1+t^4+2t^2-4t^2}{{\left(1+t^2\right)}^2}} \\ =-\sqrt{\frac{1+t^4-2t^2}{{\left(1+t^2\right)}^2}} \\ =-\sqrt{\frac{{\left(1-t^2\right)}^2}{{\left(1+t^2\right)}^2}} \\ \mathrm{Hence},\cos x=-\frac{\left(1-t^2\right)}{1+t^2} \end{gathered}$$