If $\sin x=\frac{-2 \sqrt{6}}{5}$ and $x$ lies in III quadrant, then the value of $\cot x$ is $\frac{1}{m \sqrt{6}}$. Value of $m$ is

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Application +4 | -1

\sin x=\frac{-2 \sqrt{6}}{5}

and

x

lies in III quadrant, then the value of

\cot x

\frac{1}{m \sqrt{6}}

. Value of

m

View answer and explanation

Correct option: A

Option A (6) is correct because evaluating the expression using the values of trigonometric functions at the specified angles results in 6.

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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)=