If $\;\cos \,\,{\rm{\theta }} = \frac{1}{2}\,\,\left( {x + \frac{1}{x}} \right),\,\;{\rm{then}}\,\,\;\frac{1}{2}\,\,\left( {{x^2} + \frac{1}{{{x^2}}}} \right)$ is equal to

JEE MATHEMATICS MCQ: COMPOUND ANGLES

1 Easy

1 of 1

Application +4 | -1

\;\cos \,\,\mathrm{\theta } = \frac{1}{2}\,\,\left( {x + \frac{1}{x}} \right),\,\;\mathrm{then}\,\,\;\frac{1}{2}\,\,\left( {{x^2} + \frac{1}{{{x^2}}}} \right)

is equal to

View answer and explanation

Correct option: B

\cos \,\,2\mathrm{\theta }

Option B (

\cos \,2\mathrm{\theta }

) is correct because the expression simplifies to

\cos 2\theta

using trigonometric identities and the given equation.

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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