Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x).
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LHS
= cot 4x( sin5x+sin 3x)
= cot 4 x[2 sin(5x+3x/2)/cos (5x-3x/2)]
= cot 4x(2 sin4x.cos x)
= 2 cos 4x/sin4x . sin 4x.cos x
= 2cos 4x.cos x ..(i)
RHS
= cot x(sin 5x-sin 3x)
= cot x [2 cos (5x+3x/2)sin(5x-3x/2)]
= cot x(2 cos 4x.sin x)
= 2cos x /sin x.cos 4x.sinx
= 2cos 4x.cos x ..(ii)
From eq (I) and (ii) LHS =RHS
[Formula sin a+sin b=2 sin(a+b/2).cos (a-b)] and sin a- sin b =2 cos (a+b/2).sin(a-b/2)]
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Answer-- 2 cos 4 x cosx..
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