JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

a \cos ^3 \alpha+3 a \cos \alpha \sin ^2 \alpha=m

and

a \sin ^3 \alpha+3 a \cos ^2 \alpha \sin \alpha=n

, then

(m+n)^{2 / 3}+

(m-n)^{2 / 3}

is equal to

View answer and explanation

Correct option: C

2 a^{2 / 3}

Option C (2

a^{2/3}

) is correct because, given the equations for m and n, we can find (m+n)^(2/3) + (m-n)^(2/3) by using the binomial expansion and trigonometric identities, which simplifies to 2

a^{2/3}

More Questions from TRIGONOMETRY RATIOS

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\tan \alpha=\frac{p}{q}

, where

\alpha=6 \beta

and

\alpha

being an acute angle, then

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

is equal to The value of

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

is If the

\triangle A B C

is acute angled at

C

, then The value of the expression

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

equals

\sin ^2 \frac{A}{2}+\sin ^2 \frac{B}{2}+\sin ^2 \frac{C}{2}=

\sin ^2(A / 2)+\sin ^2(B / 2)-\sin ^2(C / 2)=