JEE MATHEMATICS MCQ: COMPOUND ANGLES

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\alpha ,\;\,\beta ,\;\,\gamma \in \,\,\left[ {0,\,\;\frac{\pi }{2}} \right]

, then the value of

\frac{{\sin \,\,\left( {\alpha + \beta + \gamma } \right)}}{{\sin \,\,\alpha \,\, + \,\,\sin \,\,\beta \,\, + \,\,\sin \,\,\gamma }}

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Correct option: A

< 1

Option A (

< 1

) is correct because the sine of the sum of angles is always less than the sum of the sines of the individual angles in the given interval.

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

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\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

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\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}

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

\tan A\,\,\tan B