JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

\tan \frac{y}{2}=\sqrt{\frac{1-e}{1+e}} \tan \frac{x}{2}

, then

\frac{d y}{d x}=\ldots .

View answer and explanation

Correct option: A

\frac{\sqrt{1-e^2}}{1+e \cos x}

Option A (

\frac{{\sqrt{{1-e^2}}}}{{1+e \cos x}}

) is correct because it is the result of differentiating the given equation and simplifying using trigonometric identities.

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