JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Easy

1 of 1

Application +4 | -1

\tan \,\,\frac{{2\pi }}{5} - \tan \,\,\frac{\pi }{{15}} - \sqrt 3 \,\,\tan \,\,\frac{{2\pi }}{5}\tan \,\,\frac{\pi }{{15}}

is equal to

View answer and explanation

Correct option: D

\sqrt 3

Option D (

\sqrt 3

) is correct because the expression simplifies to

\tan(\frac{2\pi}{5} - \frac{\pi}{15}) = \tan(\frac{\pi}{3}) = \sqrt{3}

using the tangent subtraction formula.

More Questions from COMPOUND ANGLES

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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