Question #74d63

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Answer 1 · 2017-12-18T09:47:56

See explanation

#(cot(x)-tan(x))/(cot(x)+tan)#

#(cos(x)/sin(x)-sin(x)/cos(x))/(cos(x)/sin(x)+sin(x)/cos(x))#

#((cos(x)*cos(x))/(sin(x)*cos(x))-(sin(x)*sin(x))/(cos(x)*sin(x)))/((cos(x)*cos(x))/(sin(x)*cos(x))+(sin(x)*sin(x))/(cos(x)*sin(x)))#

#((cos^2(x)-sin^2(x))/(cos(x)*sin(x)))/((cos^2(x)+sin^2(x))/(cos(x)*sin(x)))#

#(cos^2(x)-sin^2(x))/(cos^2(x)+sin^2(x))#

#cos^2(x)-sin^2(x)#

Use the angle-sum identity #cos(2v)=cos^2(v)-sin^2(v)#

#cos(2x)#

Answer 2 · 2017-12-18T10:04:02

See the proof below

Reminder

#cotx=cosx/sinx#

#tanx=sinx/cosx#

#sin^2x+cos^2x=1#

#cos2x=cos^2x-sin^2x#

Therefore,

#LHS=(cotx-tanx)/(cotx+tanx)#

#=(cosx/sinx-sinx/cosx)/(cosx/sinx+sinx/cosx)#

#=((cos^2x-sin^2x)/(sinxcosx))/((cos^2x+sin^2x)/(sinxcosx))#

#=(cos^2x-sin^2x)/(cos^2x+sin^2x)#

#=cos^2x-sin^2x#

#=cos2x#

#=RHS#

#QED#