Question

Prove that:
\[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right) = \tan 3x\]

Answer 1

L.H.S = `tanx tan(pi/3 - x) tan (pi/3 + x)`

= `tanx .  (sin(pi/3 - x))/(cos(pi/3 - x)).sin(pi/3 + x)/(cos(pi/3 + x))`

= `(sinx . sin(pi/3 - x). sin(pi/3 + x))/(cosx . cos(pi/3 - x) . cos(pi/3 + x))`

= `(sinx . (sin^2  pi/3 - sin^2x))/(cosx . (cos^2  pi/3 - sin^2x))`

= `sinx/cosx((sqrt3/2)^2 - sin^2x)/((1/2)^2 - sin^2x)`

= `sinx/cosx ((3/4) - sin^2x)/((1/4) - sin^2x)`

= `sinx/cosx ((3 - 4sin^2x)/(1 - 4sin^2x))`

= `sinx/cosx ((3 - 4sin^2x)/(1 - 4(1 - cos^2x)))`

= `sinx/cosx ((3 - 4sin^2x)/(4 cos^2x - 3))`

= `(3 sinx - 4 sin^3x)/(4cos^2  -  3cosx)`

= `(sin3x)/(cos3x)`

= `tanx`