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JEE MATHEMATICS MCQ: COMPOUND ANGLES

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If

\sin ^{4} \alpha+4 \cos ^{4} \beta+2=4 \sqrt{2} \sin \alpha \cos \beta ; \alpha, \beta \in[0, \pi]

, then

\cos (\alpha+\beta)-\cos (\alpha-\beta)

is equal to:

View answer and explanation

Correct option: D

-\sqrt{2}

Option D (

-\sqrt{2}

) is correct because after applying AM-GM inequality and solving for

\alpha

and

\beta

, the expression

\cos(\alpha+\beta)-\cos(\alpha-\beta)

simplifies to

-\sqrt{2}

.

More Questions from COMPOUND ANGLES

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

\tan \left( {A - B} \right) = q

, then the value of

\tan \,\,2A

is The value of

\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)

, is
The sides of a triangle are

\sin \alpha, \cos \alpha

\sqrt{1+\sin \alpha \cos \alpha}

0<\alpha<\frac{\pi}{2}

. Then, the greatest angle of the triangle is If

\tan \,\,A = 2\,\,\tan \,\,B + \cot \,\,B,\;\;\mathrm{then}\;\,2\,\,\tan \left( {A - B} \right)\mathrm{\;}

is equal to

If

A,B,C

are angles of a triangle, then the minimum value of

{\tan ^2}\,\,\frac{A}{2} + {\tan ^2}\,\,\frac{B}{2} + {\tan ^2}\,\,\frac{C}{2}

A > 0,\,\;B > 0

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

, then the maximum value of

\tan A\,\,\tan B