Question

If $p=\frac{1}{2}si{n}^2\theta +\frac{1}{3}co{s}^2\theta$ , then

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Given,

$P=\frac{1}{2}si{n}^2\theta +\frac{1}{3}co{s}^2\theta =\frac{1}{2}(1-co{s}^2\theta )+\frac{1}{3}co{s}^2\theta =\frac{1}{2}-\frac{1}{6}co{s}^2\theta \because 0\le co{s}^2\theta \le 1\Rightarrow -\frac{1}{6}\le -\frac{1}{6}co{s}^2\theta \le 0\Rightarrow \frac{1}{3}\le \frac{1}{2}-\frac{1}{6}co{s}^2\theta \le \frac{1}{2}\Rightarrow \frac{1}{3}\le P\le \frac{1}{2}$