JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Medium

1 of 1

Application +4 | -1

A+B+C=\pi

, then

\left|\begin{array}{ccc} \sin (A+B+C) & \sin B & \cos B \\ -\sin B & 0 & \tan A \\ \cos (A+C) & -\tan A & 0 \end{array}\right|=

View answer and explanation

Correct option: A

Option A (1) is correct because the given condition

\sin x + \sin^2 x = 1

leads to

\sin x = \cos^2 x

, which simplifies the expression to 1.

More Questions from TRIGONOMETRY RATIOS

Practice All Chapter MCQs →

\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)=