Prove that tan(A+B)*tan(A-B)=cos²B-cos²A/cos²B-sin²A?

1 Answer
Swarna Islam
Dec 18, 2017

It is proved below:

Explanation:

L.H.S=tan(A+B).tan(A-B)L.H.S=tan(A+B).tan(A−B)

=(sin(A+B)sin(A-B))/(cos(A+B)cos(A-B))color(brown)[[[As,tantheta=sintheta/costheta]]

=((sinAcosB+cosAsinB)(sinAcosB-cosAsinB))/((cosAcosB-sinAsinB)(cosAcosB+sinAsinB))color(orange)[[As.sin(a+b)=sinacosb+cosasinbandcos(a+b)=cosacosb-sinasinb]]

=(sin^2Acos^2B-cos^2Asin^2B)/(cos^2Acos^2B-sin^2Asin^2B)color(blue)([As.(a-b)(a+b)=a^2-b^2])

=((1-cos^2A)cos^2B-cos^2A(1-cos^2B))/((1-sin^2A)cos^2B-sin^2A(1-cos^2B))color(green)([As.sin^2theta=1-cos^2theta andcos^2theta=1-sin^2theta])

=(cos^2B-cancel(cos^2Acos^2B)-cos^2A+cancel(cos^2Acos^2B))/(cos^2B-cancel(sin^2Acos^2B)-sin^2A+cancel(sin^2Acos^2B))

=(cos^2B-cos^2A)/(cos^2B-sin^2A)=R.H.Scolor(blue)([Proved])

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