Question

Solve the equation $tanx+tan2x+\sqrt{3}tanxtan2x=\sqrt{3}$.
Or
Prove that $sin{18}^{\circ }=\frac{\sqrt{5}-1}{4}$.

Answer 1

Given, $tanx+tan2x+\sqrt{3}tanxtan2x=\sqrt{3}$

$\Rightarrow tanx+tan2x=\sqrt{3}-\sqrt{3}tanxtan2x$

$tanx+tan2x=\sqrt{3}(1-tanx.tan2x)$

$\Rightarrow \frac{tanx+tan2x}{1-tanxtan2x}=\sqrt{3}$

$\Rightarrow tan3x=\sqrt{3}\Rightarrow tan3x=tan\frac{\pi }{3}$ $[\because tan(A+B)=\frac{tanA+tanB}{1-tanA.tanB}]$

$\Rightarrow 3x=n\pi +\frac{\pi }{3},nϵZ\Rightarrow x=\frac{n\pi }{3}+\frac{\pi }{9},nϵZ$

Or

Let $\theta ={18}^{\circ }$, then

$5\theta ={18}^{\circ }\times 5={90}^{\circ }\Rightarrow 3\theta +2\theta ={90}^{\circ }$

$\Rightarrow 2\theta ={90}^{\circ }-3\theta \Rightarrow sin2\theta =sin({90}^{\circ }-3\theta )$

$\Rightarrow sin2\theta =cos3\theta \Rightarrow 2sin\theta cos\theta =4co{s}^2\theta -3cos\theta$ $[\because cos3\theta =4co{s}^2\theta -3cos\theta ]$

$\Rightarrow 2sin\theta =4co{s}^2\theta -3\Rightarrow 2sin\theta =4-4si{n}^2\theta -3$

$\Rightarrow 4si{n}^2\theta +2sin\theta -1=0\Rightarrow sin\theta =\frac{-2\pm \sqrt{4+16}}{8}=\frac{-1\pm \sqrt{5}}{4}$

$\Rightarrow sin\theta =\frac{\sqrt{5}-1}{4}$ $[\because \theta \text{lies in 1st quadrant}\Rightarrow sin\theta >0]$

Hence, $sin{18}^{\circ }=\frac{\sqrt{5}-1}{4}$

Hence proved.