Prove that $csc 4 A + csc 8 A = cot 2 A - cot 8 A$ $csc4A+csc8A=cot2A-cot8A$ ?

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Answer 1 · 2017-10-27T10:00:28

$R H S = cot 2 A - cot 8 A$ $RHS=cot2A-cot8A$

$= \frac{cos 2 A}{sin 2 A} - \frac{cos 8 A}{sin 8 A}$ $=(cos2A)/(sin2A)-(cos8A)/(sin8A)$

$= \frac{cos 2 A sin 8 A - cos 8 A sin 2 A}{sin 2 A sin 8 A}$ $=(cos2Asin8A-cos8Asin2A)/(sin2Asin8A)$

$= \frac{sin (8 A - 2 A)}{sin 2 A sin 8 A}$ $=sin(8A-2A)/(sin2Asin8A)$

$= \frac{2 cos 2 A sin 6 A}{2 cos 2 A sin 2 A sin 8 A}$ $=(2cos2Asin6A)/(2cos2Asin2Asin8A)$

$= \frac{sin 8 A + sin 4 A}{sin 4 A sin 8 A}$ $=(sin8A+sin4A)/(sin4Asin8A)$

$= \frac{sin 8 A}{sin 4 A sin 8 A} + \frac{sin 4 A}{sin 4 A sin 8 A}$ $=(sin8A)/(sin4Asin8A)+(sin4A)/(sin4Asin8A)$

$= \frac{1}{sin 4 A} + \frac{1}{sin 8 A}$ $=1/(sin4A)+1/(sin8A)$

$= csc 4 A + csc 8 A = L H S$ $=csc4A+csc8A=LHS$

Prove that csc4A+csc8A=cot2A-cot8Acsc4A+csc8A=cot2A−cot8Acsc4A+csc8A=cot2A-cot8A?