Cambridge 3U 2c Q19 - Further Trig (1 Viewer)

[yiaghertop]

really need help with this, its so hard, if someone can show full steps that'd be a LIFESAVER
I've tried a bunch of things, full method would be awesome
given (tanx)^2=tan(x-a)tan(x-b), show that tan2x=(2sinaxsinb)/(sin(a+b))

 

[fromthethickofit]

expand tan(x-a) and tan(x-b) using the tangent subtracting identity substitute it to the given equation (tan2x=tan(x-a)tan(x-b)) then simplify the numerator and denominator then the equation will become ... and then use the double angle formula for tangent tan2x=2tanx/1-tan2x and substitute into tan2x after some algebraic manipulation.

 

[yiaghertop]

expand tan(x-a) and tan(x-b) using the tangent subtracting identity substitute it to the given equation (tan2x=tan(x-a)tan(x-b)) then simplify the numerator and denominator then the equation will become ... and then use the double angle formula for tangent tan2x=2tanx/1-tan2x and substitute into tan2x after some algebraic manipulation.

Have you used this and got the answer? Just saying this because I tried doing tan^2(x) = .... (expanding the tan(x-a)tan(x-b)) and then subbing this tan squared value into tan2x = 2tanx / 1-tan^2(x) . It ends up like a mess up of terms of tanx, tana, tanb which seems unsolvable and it doesnt go anywhere as theres no way to cancel the tanx.
If you have got the answer, could you please be a bit more detailed in the steps and algebraic manipulation? Thank you!!