Question

If tanx/tany = 1+cos2x/1+sin2x then show that

sin(3x+y) = 7sin(x—y)

Answer 1

Hey dear,

We have given,
tanx / tany = 1 + cos^2x

Using conponendo dividendo,
(tanx+tany) / (tanx-tany) = (cos^2x+sin^2x)/(2+1)
sin(x-y) / sin(x+y) = cos2x / 3
3 × sin(x-y) = sin(x+y) × cos2x

Doubling both sides,
6 sin(x-y) = 2 sin(x+y) × cos2x
6 sin(x-y) = sin(3x+y) - sin(x-y)
7 sin(x-y) = sin(3x+y)

Hope this is useful...