Question

Solve $tanx+tan2x+tan3x=tanxtan2xtan3x.$

OR

prove that $cot\frac{\pi }{24}=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}.$

Answer 1

We have $tanx+tan2x+tan3x=tanxtan2xtan3x.$

$\Rightarrow tanx+tan2x=-tan3x+tanxtan2xtan3x.$

$\Rightarrow tanx+tan2x=-tan3x(1-tanxtan2x)$

$\Rightarrow \frac{tanx+tan2x}{1-tanxtan2x}=-tan3x$

$\Rightarrow tan(x+2x)=-tan3x$

$\Rightarrow tan3x=-tan3x$

$\Rightarrow 2tan3x=0$

$\Rightarrow tan3x=0$

$\Rightarrow 3x=n\pi ,\cap \in Z$

$\therefore x=\frac{n\pi }{3},\cap \in Z$

Or

we have, $LHS=cot\left(\frac{\pi }{24}\right)$

$=\frac{cos\left(\frac{\pi }{24}\right)}{sin\left(\frac{\pi }{24}\right)}=\frac{2cos\left(\frac{\pi }{24}\right)cos\left(\frac{\pi }{24}\right)}{2sin\left(\frac{\pi }{24}\right)cos\left(\frac{\pi }{24}\right)}$

$=\frac{2co{s}^2\left(\frac{\pi }{24}\right)}{2sin\left(\frac{\pi }{24}\right)cos\left(\frac{\pi }{24}\right)}$

$=\frac{1+cos\left(\frac{\pi }{12}\right)}{sin\frac{\pi }{12}}[\because 1+cos\theta =2co{s}^2\frac{\theta }{2}]$

$=\frac{1+cos(\frac{\pi }{4}-\frac{\pi }{6})}{sin(\frac{\pi }{4}-\frac{\pi }{6})}$

$=\frac{1+cos\frac{\pi }{4}cos\frac{\pi }{6}+sin\frac{\pi }{4}sin\frac{\pi }{6}}{sin\frac{\pi }{4}cos\frac{\pi }{6}-cos\frac{\pi }{4}sincos\frac{\pi }{6}}$

$[\because cos(A-B)=cosAcosB+sinAsinBandsin(A-B)=sinAcosB-cosAsinB]$

$=\frac{1+\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}}{\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}}$

$=\frac{2\sqrt{2}+\sqrt{3}+1}{\sqrt{3}-1}$

$=\frac{(2\sqrt{2}+\sqrt{3}+1)\times (\sqrt{3}+1)}{(\sqrt{3}-1)\times (\sqrt{3}+1)}$

$=\frac{2\sqrt{6}+2\sqrt{2}+3+\sqrt{3}+\sqrt{3}+1}{3-1}$

$=\frac{2\sqrt{6}+2\sqrt{2}+2\sqrt{3}+4}{2}[\because \sqrt{4}=2]$

$=\sqrt{6}+\sqrt{2}+\sqrt{3}+2$

$=\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}=RHS$