JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

In a

\triangle {ABC}, \, {a}^2 \sin 2 {B}+{b}^2 \sin 2 {A}

is equal to (where

\Delta

is area of

\triangle {ABC}

)

View answer and explanation

Correct option: A

4 \Delta

Option A (

4 \Delta

) is correct because

a^2 \sin 2B + b^2 \sin 2A

simplifies to

4 \Delta

using trigonometric identities and the area formula.

More Questions from TRIGONOMETRY RATIOS

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

\alpha=6 \beta

\alpha

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

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

\triangle A B C

C

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

\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)=