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Trigonometrical Equations
hard
माना $S =\left\{\theta \in[-\pi, \pi]-\left\{\pm \frac{\pi}{2}\right\}: \sin \theta \tan \theta+\tan \theta=\sin 2 \theta\right\}$ है। यदि $T =\sum_{\theta \in S } \cos 2 \theta$ है, तो $T + n ( S )$ बराबर है
A
$7+\sqrt{3}$
B
$9$
C
$8+\sqrt{3}$
D
$10$
(JEE MAIN-2022)
Solution
$\sin \theta \tan \theta+\tan \theta=\sin 2 \theta$
$\tan \theta(\sin \theta+1)=\frac{2 \tan \theta}{1+\tan ^{2} \theta}$
$\tan \theta=0 \Rightarrow \theta=-\pi, 0, \pi$
$(\sin \theta+1)=2 \cdot \cos ^{2} \theta=2(1+\sin \theta)(1-\sin \theta)$
$\sin \theta=-1$ which is not possible
$\sin \theta=\frac{1}{2} \quad \theta=\frac{\pi}{6}, \frac{5 \pi}{6}$
$n ( s )=5$
$T =\cos 0+\cos 2 \pi+\cos 2 \pi+\cos \frac{\pi}{3}+\cos \frac{5 \pi}{3}$
$T =4$
$T + n ( s )=9$
Standard 11
Mathematics
