Question #74d63

2 Answers
Dec 18, 2017

See explanation

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)

Dec 18, 2017

See the proof below

Explanation:

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