JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Analyzing +4 | -1

If in a

\triangle A B C, \frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a}

, then

\angle A

is equal to

View answer and explanation

Correct option: A

90^{\circ}

Option A (

90^{\circ}

) is correct because simplifying the given equation using the sine rule and cosine rule leads to the conclusion that angle A is

90^{\circ}

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

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\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)=