In $\triangle A B C, \cos ^2 A+\cos ^2 B-\cos ^2 C=1$, then

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

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

, then

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Correct option: C

\angle C=\frac{\pi}{2}

Option C (

\angle C=\frac{\pi}{2}

) is correct because the given equation can be rearranged to show that

a^2 + b^2 = c^2

, which implies that the triangle is right-angled at C.

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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)=