Question

If ${\displaystyle \sqrt{3}}$ sin θ – cos θ = θ, then show that tan 3θ = ${\displaystyle \frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }}$

Answer 1

If ${\displaystyle \sqrt{3}}$ sin θ – cos θ = θ

To prove tan 3θ = ${\displaystyle \frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }}$

${\displaystyle \sqrt{3}}$ sin θ – cos θ = θ

${\displaystyle \sqrt{3}}$ sin θ = cos θ 

${\displaystyle \frac{\sin \theta }{\cos \theta }=\frac{1}{\sqrt{3}}}$

tan θ = tan 30°

θ = 30°

L.H.S = tan 3θ°

= tan3 (30°)

= tan 90°

= undefined (α)

R.H.S = ${\displaystyle \frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }}$

= ${\displaystyle \frac{3{\tan 30}^{\circ }-{\tan }^2{30}^{\circ }}{1-3{\tan }^2{30}^{\circ }}}$

= ${\displaystyle 3\left(\frac{1}{\sqrt{3}}\right)-{\left(\frac{1}{\sqrt{3}}\right)}^3÷1-3\times {\left(\frac{1}{\sqrt{3}}\right)}^2}$

= ${\displaystyle \sqrt{3}-\frac{1}{3\sqrt{3}}÷1-3\times \frac{1}{3}}$

= ${\displaystyle \frac{9-1}{3\sqrt{3}}÷1-1}$

= ${\displaystyle \frac{8}{3\sqrt{3}}÷0}$

= undefined (α)

∴ tan 3θ = ${\displaystyle \frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }}$