Prove that tan 225 cot 405 + cot 675 tan 765 = 0

Answers

Answer 1

Answer: $tan225^ocot405^o+cot675^otan765^o=0\n\nL=tan(180^o+45^o)cot(2*180^o+45^o)+\dots\n\dots+cot(4*180-45^o)tan(4*180^o+45^o)\n\n=tan45^ocot45^o+cot(-45^o)tan45^o=1-cot45^otan45^o\n\n=1-1=0\n\n===================================\n\ntan\alpha* cot\alpha=1\n\ncot(-\alpha)=-cot\alpha$

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Answers

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Answers

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