JEE Mains · Maths · STD 11 - Trigonometrical equations
The sum of solutions of the equation \(\frac{\cos \mathrm{x}}{1+\sin \mathrm{x}}=|\tan 2 \mathrm{x}|, \mathrm{x} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)-\left\{\frac{\pi}{4},-\frac{\pi}{4}\right\}\) is :
- A \(-\frac{11 \pi}{30}\)
- B \(\frac{\pi}{10}\)
- C \(-\frac{7 \pi}{30}\)
- D \(-\frac{\pi}{15}\)
Answer & Solution
Correct Answer
(A) \(-\frac{11 \pi}{30}\)
Step-by-step Solution
Detailed explanation
\(\frac{\cos x}{1+\sin x}=|\tan 2 x|\) \(\Rightarrow \frac{\cos ^{2} x / 2-\sin ^{2} x / 2}{(\cos x / 2+\sin x / 2)}=|\tan 2 x|\) \(\Rightarrow \tan ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan ^{2} 2 x\) \(\Rightarrow 2 x=n \pi \pm\left(\frac{\pi}{4}-\frac{x}{2}\right)\)…
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