Question

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

Answer 1

It is proved below:

Explanation:

`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.S` `color(blue)([Proved])`