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

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Answer 1 · 2018-04-19T14:07:50

#LHS=tan(pi/4 + theta) - tan(pi/4 - theta)#

#=(tan(pi/4) + tantheta)/(1-tan(pi/4)tantheta) -( tan(pi/4) -tan theta)/(1+tan(pi/4)tantheta)#

#=(1+ tantheta)/(1-tantheta) -( 1-tan theta)/(1+tantheta)#

#=((1+ tantheta)^2 -( 1-tan theta)^2)/(1-tan^2theta)#

#=(4tantheta)/(1-tan^2theta)#

# = 2tan2theta#

Answer 2 · 2018-04-19T14:30:54

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)))}#

#=2tan2theta#

#=RHS#

hence, proved! :)