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Prove that $\tan 70^{\large\circ}=2\tan 50^{\large\circ}+\tan 20^{\large\circ}$

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  • $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
  • $\tan (90-\theta)=\cot \theta$
  • $\cot \theta.\tan \theta=1$
$\tan 70^{\large\circ}=\tan(50+20)^{\large\circ}=\large\frac{\tan 50^{\large\circ}+\tan 20^{\large\circ}}{1-\tan 50^{\large\circ}\tan 20^{\large\circ}}$
$\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A\tan B}$
$\tan 70^{\large\circ}(1-\tan 50^{\large\circ}\tan 20^{\large\circ})=\tan 50^{\large\circ}+\tan 20^{\large\circ}$
$\tan 70^{\large\circ}-\tan 70^{\large\circ}\tan 50^{\large\circ}\tan 20^{\large\circ}=\tan 50^{\large\circ}+\tan 20^{\large\circ}$
$\tan 70^{\large\circ}=\tan 50^{\large\circ}+\tan 20^{\large\circ}+\tan 70^{\large\circ}\tan 50^{\large\circ}\tan 20^{\large\circ}$
$\Rightarrow \tan 50^{\large\circ}+\tan 20^{\large\circ}+\tan (90^{\large\circ}-20^{\large\circ})\tan 50^{\large\circ}\tan 20^{\large\circ}$
$\Rightarrow \tan 50^{\large\circ}+\tan 20^{\large\circ}+\cot 20^{\large\circ}\tan 50^{\large\circ}\tan 20^{\large\circ}$
$\tan (90-\theta)=\cot \theta$
$\cot \theta.\tan \theta=1$
$\Rightarrow \tan 50^{\large\circ}+\tan 20^{\large\circ}+\tan 50^{\large\circ}$
$\Rightarrow 2\tan 50^{\large\circ}+\tan 20^{\large\circ}$=RHS
Hence proved.
answered May 21, 2014 by sreemathi.v

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