JEE MATHEMATICS MCQ: MULTIPLE AND SUBMULTIPLE ANGLES

1 Medium

1 of 1

Application +4 | -1

0 < A < \frac{\pi }{6}

and

\sin A + \cos A = \frac{{\sqrt 7 }}{2}

, then

\tan \frac{A}{2} =

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the solution for

\tan \frac{A}{2}

derived from the given trigonometric equation and the constraint on A.

More Questions from MULTIPLE AND SUBMULTIPLE ANGLES

Practice All Chapter MCQs →

1 + \cos \,\,x = k

x

\sin \,\,\frac{x}{2}

\sin 12^\circ \cdot \sin 48^\circ \cdot \sin 54^\circ =

\alpha ,\;\,\,\beta \left( {\alpha \ne \beta } \right)

a\,\,\cos \,\,\theta + b\,\,\sin \,\,\theta = c

\tan \,\,\left( {\frac{{\alpha + \beta }}{2}} \right)

{\cos ^4}\theta - {\sin ^4}\theta

\cos ^{2} 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^{2} 50^{\circ}

\triangle P Q R, \angle R=\frac{\pi}{2}

\tan \left(\frac{P}{2}\right)

\tan \left(\frac{Q}{2}\right)

a x^2+b x+c=0, a \neq 0