# How do you prove tan(pi/4 + theta) - tan(pi/4 - theta) = 2tan2theta?

##### 2 Answers
Apr 19, 2018

$L H S = \tan \left(\frac{\pi}{4} + \theta\right) - \tan \left(\frac{\pi}{4} - \theta\right)$

$= \frac{\tan \left(\frac{\pi}{4}\right) + \tan \theta}{1 - \tan \left(\frac{\pi}{4}\right) \tan \theta} - \frac{\tan \left(\frac{\pi}{4}\right) - \tan \theta}{1 + \tan \left(\frac{\pi}{4}\right) \tan \theta}$

$= \frac{1 + \tan \theta}{1 - \tan \theta} - \frac{1 - \tan \theta}{1 + \tan \theta}$

$= \frac{{\left(1 + \tan \theta\right)}^{2} - {\left(1 - \tan \theta\right)}^{2}}{1 - {\tan}^{2} \theta}$

$= \frac{4 \tan \theta}{1 - {\tan}^{2} \theta}$

$= 2 \tan 2 \theta$

Apr 19, 2018

We know,

color(blue)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB)=> tan(A±B)*{1∓tanAtanB}=tanA±tanB)

So,

(tanunderbrace((π/4+theta))_color(blue)text(A)-tanunderbrace((π/4-theta)_color(blue)text(B)))

= tan(cancel(π/4)+theta-cancel(π/4)+theta)*{1+tan(π/4+theta) tan (π/4-theta)}

Again applying, color(blue)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB)=> tan(A±B)*1∓tanAtanB=tanA±tanB)

=tan(2theta){1+(tan(π/4)+tantheta)/(1-tan(π/4)tantheta)×(tan(π/4)-tantheta)/(1+tan(π/4)tantheta)}

We know,

tan(π/4)=1

So,

=tan(2theta){1+cancel((1+tantheta))/(cancel((1-tantheta)))×(cancel((1-tantheta)))/(cancel((1+tantheta)))}

$= 2 \tan 2 \theta$

$= R H S$

hence, proved! :)