Question

Prove that cot(A+15)-tan(A-15) = (4cos2A) / (1+2sin2A)

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Answer 1

$cot(A+15) - tan(A-15) \\=\frac{cos(A+15)}{sin(A+15)} - \frac{sin(A-15)}{ cos(A - 15) } \\\\=\frac{cos(A+15)cos(A-15) - sin(A-15)sin(A+15)}{ sin(A+15)cos(A-15)}\\ \\=\frac{cos2A }{sin(A+15)cos(A-15) }\\= \frac{2cos2A}{ [sin2A +sin30] }\\ =\frac{2cos2A}{ [ sin2A + 1/2] }\\=\frac{2cos2A}{ [(2sin2A+1)/2] }\\=\frac{4cos2A}{(2sin2A + 1)}$

$LHS=RHS\\Hence,Proved.$