sum solving......

prove ... cot4x(sin5x+sin3x) = cotx(sin5x-sin3x)

3 Answers

1
aditya ravichandran ·

L.H.S=\frac{\cos(4x)}{\sin(4x)}(2\sin(4x).\cos(x))=2\cos(x)\cos(4x) \\ R.H.S=\frac{\cos(x)}{\sin(x)}(2\cos(4x).\sin(x))=2\cos(x)\cos(4x) \\ L.H.S=R.H.S

21
Arnab Kundu ·

LHS=cot(4x)[sin(5x)+sin(3x)] =cot(4x)[2sin(4x)cos(x)] =2\frac{cos(4x)}{sin(4x)}sin(4x)cos(x) =2cos(4x)cos(x) =2cos(4x)cos(x)\frac{sin(x)}{sin(x)} =2cos(4x)\frac{cos(x)}{sin(x)}sin(x) =2cos(4x)cot(x)sin(x) =cot(x)[2cos(4x)sin(x)] =cotx[sin(5x)-cos(3x)]

21
Arnab Kundu ·

please ignore the <br>s.........i dont know why they are coming

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