JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Application +4 | -1

\tan \alpha=\frac{b}{a}, a > b > 0

and

0 < \alpha < \frac{\pi}{4}

, then

\sqrt{\frac{a+b}{a-b}}-\sqrt{\frac{a-b}{a+b}}

is equal to

View answer and explanation

Correct option: A

\frac{2 \sin \alpha}{\sqrt{\cos 2 \alpha}}

Option A (

\frac{{2 \sin \alpha}}{{\sqrt{{\cos 2 \alpha}}}}

) is correct because it is derived by simplifying the given expression using trigonometric identities and the given condition

\tan \alpha = \frac{{b}}{{a}}

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