Prove that `sinthetasec3theta=1/2[tan3theta-tantheta]` and hence find the sum to `n` terms of
62 views · Published 13 October 2018 · 6:05 · Indexed 25 September 2026
Channel: Doubtnut · 2018 · Education
To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW Prove that `sinthetasec3theta=1/2[tan3theta-tantheta]` and hence find the sum to `n` terms of the series `sinthetasec3theta+sin3thetasec3^2theta+sin3^2thetasec3^3theta+....`
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