JEE MATHEMATICS MCQ: MULTIPLE AND SUBMULTIPLE ANGLES

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Application +4 | -1

0<\alpha<\pi / 6

and

\sin \alpha+\cos \alpha=\sqrt{7} / 2

, then

\tan \alpha / 2

is equal to

View answer and explanation

Correct option: A

\frac{\sqrt{7}-2}{3}

Option A (

\frac{{\sqrt{{7}}-2}}{{3}}

) is correct because solving the trigonometric equation and considering the given range for

\alpha

, the value of

\tan \frac{{\alpha}}{{2}}

\frac{{\sqrt{{7}}-2}}{{3}}

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