How to prove that #tan(pi/4+alpha)-tan(pi/4-alpha)=2tan2alpha#?
2 Answers
Please see the proof below
Explanation:
We need
Therefore,
Explanation:
#"using the "color(blue)"trigonometric identities"#
#•color(white)(x)tan(x+-y)=(tanx+-tany)/(1∓tanxtany)#
#•color(white)(x)tan2x=(2tanx)/(1-tan^2x)#
#"consider the left side"#
#(tan(pi/4)+tanalpha)/(1-tan(pi/4)tanalpha)-(tan(pi/4)-tanalpha)/(1+tan(pi/4)tanalpha)#
#=(1+tanalpha)/(1-tanalpha)-(1-tanalpha)/(1+tanalpha)#
#=((1+tanalpha)^2-(1-tanalpha)^2)/((1-tanalpha)(1+tanalpha))#
#=(1+2tanalpha+tan^2alpha-1+2tanalpha-tan^2alpha)/(1-tan^2alpha)#
#=(4tanalpha)/(1-tan^2alpha)#
#=(2tanalpha+2tanalpha)/(1-tan^2alpha)#
#=(2tanalpha)/(1-tan^2alpha)+(2tanalpha)/(1-tan^2alpha)#
#=tan2alpha+tan2alpha#
#=2tan2alpha="right side "rArr"verified"#