JEE Mains · Maths · STD 11 - Trigonometrical equations
Let \(S=\{\theta \in[0,2 \pi): \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0\}\). Then \(\sum_{\theta \in S } \sin ^2\left(\theta+\frac{\pi}{4}\right)\) is equal to
- A \(4\)
- B \(6\)
- C \(8\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(\tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0\) \(\tan (\pi \cos \theta)=-\tan (\pi \sin \theta)\) \(\tan (\pi \cos \theta)=\tan (-\pi \sin \theta)\) \(\pi \cos \theta=n \pi-\pi \sin \theta\) \(\sin \theta+\cos \theta= n \text { where } n \in I\) possible values are…
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