If $P = \frac{1}{2}\,\,{\sin ^2}\theta \,\, + \,\,\frac{1}{3}\,\,{\cos ^2}\theta $, then

$ - \frac{{\sqrt {13} }}{6} \le P \le \frac{{\sqrt {13} }}{6}$

JEE MATHEMATICS MCQ: TRIGONOMETRY RATIOS

1 Hard

1 of 1

Application +4 | -1

P = \frac{1}{2}\,\,{\sin ^2}\theta \,\, + \,\,\frac{1}{3}\,\,{\cos ^2}\theta

, then

View answer and explanation

Correct option: A

\frac{1}{3} \le P \le \frac{1}{2}

Option A (

\frac{1}{3} \le P \le \frac{1}{2}

) is correct because by rewriting P in terms of cos²θ and using the range of cos²θ, we can find the range of P.

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)=