If ${\rm{\;}}\tan \,\,{\rm{\alpha }} = \frac{1}{{\sqrt {x\left( {{x^2} + x + 1} \right)} }},\,\,\tan \,\,{\rm{\beta }} = \frac{{\sqrt x }}{{\sqrt {{x^2} + x + 1} }}{\rm{\;}}$ and ${\rm{\;}}\tan \,\,{\rm{\gamma }} = \sqrt {{x^{ - 3}} + {x^{ - 2}} + {x^{ - 1}},} $ then ${\rm{\;\alpha }} + {\rm{\beta \;}}$ is

JEE MATHEMATICS MCQ: COMPOUND ANGLES

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\mathrm{\;}\tan \,\,\mathrm{\alpha } = \frac{1}{{\sqrt {x\left( {{x^2} + x + 1} \right)} }},\,\,\tan \,\,\mathrm{\beta } = \frac{{\sqrt x }}{{\sqrt {{x^2} + x + 1} }}\mathrm{\;}

and

\mathrm{\;}\tan \,\,\mathrm{\gamma } = \sqrt {{x^{ - 3}} + {x^{ - 2}} + {x^{ - 1}},}

then

\mathrm{\;\alpha } + \mathrm{\beta \;}

View answer and explanation

Correct option: A

\gamma

Option A (

\gamma

) is correct because by using the tangent addition formula and simplifying,

\tan(\alpha + \beta)

equals

\tan(\gamma)

, thus

\alpha + \beta = \gamma

Practice All Chapter MCQs →

\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