JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

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Analyzing +4 | -1

The range of

\cos \theta\left(\sin \theta+\sqrt{\sin ^2 \theta+\sin ^2 \alpha}\right)

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Correct option: A

\left[-\sqrt{1+\sin ^2 \alpha}, \sqrt{1+\sin ^2 \alpha}\right]

Option A is correct because the range is found by solving the trigonometric equation and considering the discriminant of the resulting quadratic equation in terms of tan θ.

More Questions from TRIGONOMETRY RATIOS

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\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 The value of

\cos 12^{\circ}+\cos 84^{\circ}+\cos 156^{\circ}+\cos 132^{\circ}

is If the

\triangle A B C

is acute angled at

C

, then The value of the expression

{\sin ^6}\,\,\theta + {\cos ^6}\,\,\theta + 3{\sin ^2}\,\,\theta \,\,{\cos ^2}\,\,\theta

equals

\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)=