JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Medium

1 of 1

Analyzing +4 | -1

\tan (\alpha-\beta)=\frac{\sin 2 \beta}{3-\cos 2 \beta}

, then

\tan \alpha=f(\beta)

. The value of

f\left(\frac{\pi}{4}\right)

equals

View answer and explanation

Correct option: A

Option A (2) is correct because substituting

\beta = \frac{\pi}{4}

into the given equation and solving for

\tan \alpha

yields 2.

More Questions from COMPOUND ANGLES

Practice All Chapter MCQs →

\tan \left( {A + B} \right) = p

\tan \left( {A - B} \right) = q

\tan \,\,2A

\cos \,\,x\,\,\cos \,\,y\,\,\sin \,\,\left( {x - y} \right) + \cos \,\,y\,\,\cos \,\,z\,\,\sin \,\,\left( {y - z} \right)

+ \cos \,\,z\,\,\cos \,\,x\,\,\sin \,\,\left( {z - x} \right) + \sin \,\,\left( {x - y} \right)\sin \,\,\left( {y - z} \right)\,\,\sin \,\,\left( {z - x} \right)

\sin \alpha, \cos \alpha

\sqrt{1+\sin \alpha \cos \alpha}

0<\alpha<\frac{\pi}{2}

\tan \,\,A = 2\,\,\tan \,\,B + \cot \,\,B,\;\;\mathrm{then}\;\,2\,\,\tan \left( {A - B} \right)\mathrm{\;}

A,B,C

{\tan ^2}\,\,\frac{A}{2} + {\tan ^2}\,\,\frac{B}{2} + {\tan ^2}\,\,\frac{C}{2}

A > 0,\,\;B > 0

A + B = \frac{\pi }{3}

\tan A\,\,\tan B