If $\cos \left( {A - B} \right) = \frac{3}{5}$ and $\tan A\,\,\tan B = 2$, then

$\sin A\,\,\sin B = - \frac{2}{5}$

$\cos \,\,\left( {A + B} \right) = - \frac{1}{5}$

JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Easy

1 of 1

Application +4 | -1

\cos \left( {A - B} \right) = \frac{3}{5}

and

\tan A\,\,\tan B = 2

, then

View answer and explanation

Correct option: A

\cos A\,\,\cos B = \frac{1}{5}

Option A (cos A cos B = 1/5) is correct because, given the conditions, solving for tan A and tan B allows us to find cos A cos B = 1/5.

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\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