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Mathematics Question on Trigonometric Functions

If sin θ=2t1+t2\theta = \frac{2t}{1+t^{2}} and ?? lies in the second quadrant, then cos θcos\,\theta is equal to

A

1−t21+t2\frac{1-t^{2}}{1+t^{2}}

B

t2−11+t2\frac{t^{2}-1}{1+t^{2}}

C

−∣1−t2∣1+t2\frac{-\left|1-t^{2}\right|}{1+t^{2}}

D

1+t2∣1−t2∣\frac{1+t^{2}}{\left|1-t^{2}\right|}

Answer

−∣1−t2∣1+t2\frac{-\left|1-t^{2}\right|}{1+t^{2}}

Explanation

Solution

Given, sin⁡θ=2t1+t2\sin \theta=\frac{2 t}{1+t^{2}}
Since, cos⁡θ\cos \theta lies in the second quadrant.
∴cos⁡θ<0\therefore \cos \theta < 0
⇒cos⁡θ=−1−sin⁡2θ\Rightarrow \cos \theta =-\sqrt{1-\sin ^{2} \theta}
=−1−4t2(1+t2)2=-\sqrt{1-\frac{4 t^{2}}{\left(1+t^{2}\right)^{2}}}
=−(1−t2)2(1+t2)2=-\sqrt{\frac{\left(1-t^{2}\right)^{2}}{\left(1+t^{2}\right)^{2}}}
=−∣1−t2∣1+t2=-\frac{\left|1-t^{2}\right|}{1+t^{2}}