If $\tan x=\frac{b}{a}$, then $\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=$

$\frac{2 \sin x}{\sqrt{\sin 2 x}}$

$\frac{2 \cos x}{\sqrt{\sin 2 x}}$

$\frac{2 \sin x}{\sqrt{\cos 2 x}}$

JEE MATHEMATICS MCQ: MULTIPLE AND SUBMULTIPLE ANGLES

1 Easy

1 of 1

Application +4 | -1

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

, then

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

View answer and explanation

Correct option: B

\frac{2 \cos x}{\sqrt{\cos 2 x}}

Option B (

\frac{{2 \cos x}}{{\sqrt{{\cos 2 x}}}}

) is correct because, after using componendo and dividendo and simplifying, the expression equals

\frac{{2 \cos x}}{{\sqrt{{\cos 2 x}}}}

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