If k 1 = tan 27θ – tan θ and k _ 2 = 3 + 3 9 + 9 27 , then
Mathematics · JEE Advanced · NTA Exams — Trigonometry
If k1 = tan 27θ – tan θ and
\(\mathrm{k}_{2}=\frac{\sin \theta}{\cos 3 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin 9 \theta}{\cos 27 \theta}, \text { then }\)
\(\mathrm{k}_{2}=\frac{\sin \theta}{\cos 3 \theta}+\frac{\sin 3 \theta}{\cos 9 \theta}+\frac{\sin 9 \theta}{\cos 27 \theta}, \text { then }\)
- k1 = 2k2
- k1 = k2 + 4
- k1 = k2
- none of these
Answer
(A) k 1 = 2k 2
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