JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Analyzing +4 | -1

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

, where

\alpha=6 \beta

and

\alpha

being an acute angle, then

\frac{1}{2}[p \operatorname{cosec} 2 \beta-q \sec 2 \beta]

is equal to

View answer and explanation

Correct option: A

\sqrt{p^2+q^2}

Option A (

\sqrt{{p^2 + q^2}}

) is correct because the expression simplifies to this value using trigonometric identities and the given relation

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

More Questions from TRIGONOMETRY RATIOS

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\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)=

\frac{\sin 2 A}{1+\cos 2 A} \cdot \frac{\cos A}{1+\cos A}=