JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Analyzing +4 | -1

In a

\triangle A B C, \cos B \cdot \cos C+\sin B \cdot \sin C \cdot \sin ^2 A=1

Then, the triangle is

View answer and explanation

Correct option: A

right angled isosceles

Option A (right angled isosceles) is correct because the given equation implies that

B=C

and

A = \frac{\pi}{2}

, making the triangle right-angled and isosceles.

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)=