Question

If $tanx+tan(x+\frac{\pi }{3})+tan(x+\frac{2\pi }{3})=3,$ prove that $\frac{3tanx-ta{n}^{3}x}{1-3ta{n}^{2}x=1.}$

Or

If $sin\theta =nsin(\theta +2\alpha ),$ prove that $tan(\theta +\alpha )=\frac{1+n}{1-n}tan\alpha .$

Answer 1

we have ,$tanx+tan(x+\frac{\pi }{3})+tan(x+\frac{2\pi }{3})=3$

$\Rightarrow tanx+tan(x+\frac{\pi }{3})+tan(x+\pi -\frac{\pi }{3})=3$

$\Rightarrow tanx+tan(x+\frac{\pi }{3})+tan\{\pi +(x-\frac{\pi }{3})\}=3$

$\Rightarrow tanx+tan(x+\frac{\pi }{3})+(x-\frac{\pi }{3})=3$

$[\because tan(\pi +\theta )=tan\theta ]$ $\Rightarrow tanx+\frac{tanx+tan\frac{\pi }{3}}{1-tanx.tan\frac{\pi }{3}}+\frac{tanx-tan\frac{\pi }{3}}{1+tanx.\frac{\pi }{3}}$

$[\because tan(A+B)=\frac{tanA+taB}{1-tanAtanB}andtan(A-B)=\frac{tanA-tanB}{1+tanAtanB}]$

$\Rightarrow \frac{[tanx-3ta{n}^{3}x+tanx+\sqrt{3}ta{n}^{2}x+\sqrt{3}+3tanx+tanx\sqrt{3}ta{n}^{2}x-\sqrt{3}+3tanx]}{1-3ta{n}^{2}x}=3$

$\Rightarrow \frac{9tanx-ta{n}^{3}x}{1-3ta{n}^{2}x}=3\Rightarrow \frac{3tanx-ta{n}^{3}x}{1-3ta{n}^{2}x}=1$

Hence proved

Or

We have, $sin\theta =nsin(\theta +2\alpha )$

$\Rightarrow \frac{sin(\theta +2\alpha )}{sin\theta }=\frac{1}{n}$

$\Rightarrow \frac{sin(\theta +2\alpha )+sin\theta }{sin\theta }=\frac{1+n}{1-n}$

[using componendo and dividendo rule]

$\Rightarrow \frac{2sin\frac{(\theta +2\alpha +\theta )}{2}cos\frac{(\theta +2\alpha -\theta )}{2}}{2cos\frac{(\theta +2\alpha +\theta )}{2}sin\frac{(\theta +2\alpha -\theta )}{2}}=\frac{1+n}{1-n}$

$[\because sinC+sinD=2sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)andsinC-sinD=2cos\left(\frac{C+D}{2}\right)sin\left(\frac{C-D}{2}\right)]$

$\Rightarrow \frac{2sin(\theta +\alpha )cos\alpha }{2cos(\theta +\alpha )sin\alpha }=\frac{1+n}{1-n}$

$\Rightarrow tan(\theta +\alpha )cot\alpha =\frac{1+n}{1-n}$

$\therefore tan(\theta +\alpha )=\frac{1+n}{1-n}tan\alpha Henceproved.$