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