Question

For $\frac{-\pi }{2}$ < $\theta$ < $\frac{\pi }{2}$ , $\frac{sin\theta +sin2\theta }{1+cos\theta +cos2\theta }$ lies in the interval.

• A. (,)
• B. (-2, 2)
• C. (0,)
• D. (-1, 1)

Answer 1

The correct option is A

(,)

We will first try to simplify the expression and find the range.

$\frac{sin\theta +sin2\theta }{1+cos\theta +cos2\theta }$
= $\frac{sin\theta +2sin\theta cos\theta }{2co{s}^2\theta +cos\theta }$

= $\frac{sin\theta }{cos\theta }$ $\left(\frac{1+2cos\theta }{1+2cos\theta }\right)$

Since, $cos\theta$ is +ve in $\frac{-\pi }{2}$, $\frac{\pi }{2}$

We can cancel 1 + 2 $cos\theta$.

$\Rightarrow$ $\frac{sin\theta }{cos\theta }$ = $\tan \theta$.We know $\tan \theta$ varies from $-\infty$ to $+\infty$ as $\theta$ goes from $\frac{-\pi }{2}$ to $\frac{\pi }{2}$.

$\Rightarrow$ Range = ($-\infty$,$\infty$)

Key steps: (1) Simplifying the expression

(2) Range of tan$\theta$

(3) cos$\theta$ is +ve in $(\frac{\pi }{2},\frac{\pi }{2})$