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How can we prove $\operatorname{cosec}(2A) + \operatorname{cosec}(4A) + \operatorname{cosec}(8A) = \cot(A) - \cot(8A)$?

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Put the right side to the left. $$cosec(2A)+cosec(4A)+cosec(8A)-cot(A)+cot(8A)=0$$ Then express the left side by sines and cosines. You get: $$\frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{1}{sin(8A)}-\frac{cos(A)}{sin(A)}+\frac{cos(8A)}{sin(8A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{1+cos(8A)}{sin(8A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{2cos^2(4A)}{2sin(4A)cos(4A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{cos(4A)}{sin(4A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{1+cos(4A)}{sin(4A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{2cos^2(2A)}{2sin(2A)cos(2A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1}{sin(2A)}+\frac{cos(2A)}{sin(2A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{1+cos(2A)}{sin(2A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{2cos^2(A)}{2sin(A)cos(A)}-\frac{cos(A)}{sin(A)} = $$ $$= \frac{cos(A)}{sin(A)}-\frac{cos(A)}{sin(A)} = 0$$

Q.E.D.

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  • $\begingroup$ I think a hint would have been better, since the OP did not show any context in their question. $\endgroup$ – Toby Mak May 19 at 12:55

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