JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

\sin \theta=\frac{2 t}{1+t^2}

and

\theta

lies in second quadrant, then

\cos \theta

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because, given

\sin \theta = \frac{{2t}}{{1+t^2}}

and

\theta

is in the second quadrant,

\cos \theta

must be negative, leading to the expression

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

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