How do you prove tan(pi/4 + theta) - tan(pi/4 - theta) = 2tan2thetatan(π4+θ)−tan(π4−θ)=2tan2θ?

2 Answers
P dilip_k
Apr 19, 2018

LHS=tan(pi/4 + theta) - tan(pi/4 - theta)LHS=tan(π4+θ)−tan(π4−θ)

=(tan(pi/4) + tantheta)/(1-tan(pi/4)tantheta) -( tan(pi/4) -tan theta)/(1+tan(pi/4)tantheta)=tan(π4)+tanθ1−tan(π4)tanθ−tan(π4)−tanθ1+tan(π4)tanθ

=(1+ tantheta)/(1-tantheta) -( 1-tan theta)/(1+tantheta)=1+tanθ1−tanθ−1−tanθ1+tanθ

=((1+ tantheta)^2 -( 1-tan theta)^2)/(1-tan^2theta)=(1+tanθ)2−(1−tanθ)21−tan2θ

=(4tantheta)/(1-tan^2theta)=4tanθ1−tan2θ

= 2tan2theta=2tan2θ

Kalit Gautam
Apr 19, 2018

We know,

color(blue)(tan(A±B)=(tanA±tanB)/(1∓tanAtanB)=> tan(A±B)*{1∓tanAtanB}=tanA±tanB)tan(A±B)=tanA±tanB1∓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)))}

=2tan2theta

=RHS

hence, proved! :)

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