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

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`

Prove that Tan X Tan ( π 3 − X ) Tan ( π 3 + X ) = Tan 3 X - Mathematics

Question

Prove that
$\tan x \tan (\frac{π}{3} - x) \tan (\frac{π}{3} + x) = \tan 3 x$

Sum

Solution

L.H.S = $\tan x \tan (\frac{π}{3} - x) \tan (\frac{π}{3} + x)$

= $\tan x . \frac{\sin (\frac{π}{3} - x)}{\cos (\frac{π}{3} - x)} . \frac{\sin (\frac{π}{3} + x)}{\cos (\frac{π}{3} + x)}$

= $\frac{\sin x . \sin (\frac{π}{3} - x) . \sin (\frac{π}{3} + x)}{\cos x . \cos (\frac{π}{3} - x) . \cos (\frac{π}{3} + x)}$

= $\frac{\sin x . (\sin^{2} \frac{π}{3} - \sin^{2} x)}{\cos x . (\cos^{2} \frac{π}{3} - \sin^{2} x)}$

= $\frac{\sin x}{\cos x} \frac{{(\frac{\sqrt{3}}{2})}^{2} - \sin^{2} x}{{(\frac{1}{2})}^{2} - \sin^{2} x}$

= $\frac{\sin x}{\cos x} \frac{(\frac{3}{4}) - \sin^{2} x}{(\frac{1}{4}) - \sin^{2} x}$

= $\frac{\sin x}{\cos x} (\frac{3 - 4 \sin^{2} x}{1 - 4 \sin^{2} x})$

= $\frac{\sin x}{\cos x} (\frac{3 - 4 \sin^{2} x}{1 - 4 (1 - \cos^{2} x)})$

= $\frac{\sin x}{\cos x} (\frac{3 - 4 \sin^{2} x}{4 \cos^{2} x - 3})$

= $\frac{3 \sin x - 4 \sin^{3} x}{4 \cos^{2} - 3 \cos x}$

= $\frac{\sin 3 x}{\cos 3 x}$

= $\tan x$

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Transformation Formulae

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Q 7 Q 6.2 Q 8

Chapter 8: Transformation formulae - Exercise 8.1 [Page 7]

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Chapter 8 Transformation formulae
Exercise 8.1 | Q 7 | Page 7

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