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Trigonometrical Equations
hard
ધારોકે $S=\{\theta \in[0,2 \pi): \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0\} .$ તો $\sum_{\theta \in s} \sin ^2\left(\theta+\frac{\pi}{4}\right)=...........$.
A
$4$
B
$6$
C
$8$
D
$2$
(JEE MAIN-2023)
Solution
$\tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0$
$\tan (\pi \cos \theta)=-\tan (\pi \sin \theta)$
$\tan (\pi \cos \theta)=\tan (-\pi \sin \theta)$
$\pi \cos \theta=n \pi-\pi \sin \theta$
$\sin \theta+\cos \theta= n \text { where } n \in I$
possible values are $n =0,1$ and $-1$ because
$-\sqrt{2} \leq \sin \theta+\cos \theta \leq \sqrt{2}$
Now it gives $\theta \in\left\{0, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{3 \pi}{2}, \pi\right\}$
So $\sum_{\theta \in S} \sin ^2\left(\theta+\frac{\pi}{4}\right)=2(0)+4\left(\frac{1}{2}\right)=2$
Standard 11
Mathematics
