JEE MATHEMATICS MCQ: COMPOUND ANGLES

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Analyzing +4 | -1

\frac{{\tan \,\,\alpha + \tan \,\,\beta }}{{\cot \,\,\alpha + \cot \,\,\beta }}\,\, + \,\,{\left\{ {\cos \,\,\left( {\alpha - \beta } \right)\,\,\sec \,\,\left( {\alpha + \beta } \right) + 1} \right\}^{ - 1}} = 1

, then tan⁡α

\tan \,\,\alpha \,\,\tan \,\,\beta

is equal to

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Correct option: A

The correct answer is A (1) because simplifying the given trigonometric expression leads to

\tan \alpha \tan \beta = 1

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