Question

Show that : Sin 3x / sin x - Cos 3x / cos x = 2​

Answer 1

Answer:

LHS:- Sin3x /Sin x - Cos 3x/Cosx

Substituting Sin 3x = 3 Sin x - 4 Sin^3 x and Cos 3x = 4 Cos^3 x - 3 Cosx in place of numerators,

=> [ (3 Sinx - 4 Sin^3 x ) / Sinx ] - [ (4 Cos^3 x - 3 Cosx) / Cosx ]

Taking common factors Sinx and Cosx out of the equation,

[ Sinx (3 - 4 Sin^2 x) / Sinx ] - [ Cosx (4 Cos^2 x - 3) / Cosx ]

Canceling common factors,

=> 3 - 4 Sin^2 x - (4 Cos^2 x -3 )

=> 3 - 4 Sin ^2 x - 4 Cos^2 x + 3

=> 6 - 4( Sin^2 x + Cos ^2 X)

Since Sin ^2 x + Cos ^2 x = 1,

=> 6 - 4 (1)

=> 2 = RHS

Answer 2

Answer:

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