JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Easy

1 of 1

Application +4 | -1

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

and

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

, then the value of

\tan \,\,2A

View answer and explanation

Correct option: D

\frac{{p\,\, + \,\,q}}{{1\,\, - \,\,pq}}

Option D (

\frac{{p\\ + \\ q}}{{1\\ - \\ pq}}

) is correct because using the tangent addition formula,

\tan 2A

can be expressed in terms of p and q as shown.

More Questions from COMPOUND ANGLES

Practice All Chapter MCQs →

\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

\alpha + \beta + \gamma = 2\pi