JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Medium

1 of 1

Analyzing +4 | -1

x + y - z + xyz = 0

, then

\frac{{2x}}{{1 - {x^2}}} + \frac{{2y}}{{1 - {y^2}}} - \frac{{2z}}{{1 - {z^2}}}

is equal to

View answer and explanation

Correct option: C

\frac{8 x y z}{\left[\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)\right]}

Option D (

\frac{{-8 x y z}}{{\left[(1-x^2)(1-y^2)(1-z^2)\right]}}

) is the correct answer because it is derived using trigonometric substitutions and the tangent addition formula.

More Questions from COMPOUND ANGLES

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