Let $$S = \left\{ {\theta \in [ - \pi ,\pi ] - \left\{ { \pm \,\,{\pi \over 2}} \right\}:\sin \theta \tan \theta + \tan \theta = \sin 2\theta } \right\}$$.
If $T = \sum\limits_{\theta \, \in \,S}^{} {\cos 2\theta }$, then T + n(S) is equal to :
Solution
$\tan \theta(\sin \theta+1)-\sin 2 \theta=0$
<br/><br/>
$$
\begin{aligned}
&\tan \theta\left(\sin \theta+1-2 \cos ^{2} \theta\right)=0 \\\\
&\Rightarrow \tan \theta=0 \text { or } 2 \sin ^{2} \theta+\sin \theta-1=0 \\\\
&\Rightarrow(2 \sin \theta+1)(\sin \theta-1)=0 \\\\
&\Rightarrow \sin \theta=\frac{-1}{2} \text { or } 1
\end{aligned}
$$
<br/><br/>
But, $\sin \theta=1$ not possible
<br/><br/>
$\theta=0, \pi,-\pi,-\frac{\pi}{6}, \frac{-5 \pi}{6}$
<br/><br/>
$\mathrm{n}(\mathrm{S})=5$
<br/><br/>
$$
T=\sum \cos 2 \theta=\cos 0^{\circ}+\cos 2 \pi+\cos (-2 \pi)
+\cos \left(-\frac{5 \pi}{3}\right)+\cos \left(-\frac{\pi}{3}\right)
$$<br/><br/>= 4
About this question
Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities
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