Question

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

Prove that $1+cos2x+cos4x+cos6x=4cosxcos2xcos3x$

Answer 1

We know that,

$tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$

$\Rightarrow tan({70}^{\circ }-{20}^{\circ })=\frac{tan{70}^{\circ }-tan{20}^{\circ }}{1+tan{70}^{\circ }tan{20}^{\circ }}[\because A={70}^{\circ },B={20}^{\circ }]$

$\Rightarrow tan{50}^{\circ }=\frac{tan{70}^{\circ }-tan{20}^{\circ }}{1+tan{70}^{\circ }cot{70}^{\circ }}[\because tan{20}^{\circ }=cot({90}^{\circ }-{20}^{\circ }\left)\right]$

$\Rightarrow tan{50}^{\circ }=\frac{tan{70}^{\circ }-tan{20}^{\circ }}{1+1}[\because tan\theta .cot\theta =1]$

$\Rightarrow 2tan{50}^{\circ }=tan{70}^{\circ }-tan{20}^{\circ }$

$\therefore tan{70}^{\circ }=tan{20}^{\circ }+2tan{50}^{\circ }$

Or

We have,

LHS$=(1+cos2x)+(cos4x+cos6x)$

$=2co{s}^{2}x+2cos\frac{4x+6x}{2}.cos\frac{4x-6x}{2}$

$=2co{s}^{2}x+2cos5x.cosx[\because cos(-\theta )=cos\theta ]$

$[\because 1+cos2x=2co{s}^{2}xandcosx+cosy=2cos\left(\frac{x+y}{2}cos\left(\frac{x-y}{2}\right)\right)]$

$=2cosx(cosx+cos5x)=cosx(2cos\frac{x+5x}{2}.cos\frac{x-5x}{2})$

$=2cosx(2cos3x.cos2x)$

$=4cosxcos2xcos3x$=RHS