Question

If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.

Answer 1

Here,

m² − n² = (tan A + sin A)² − (tan A − sin A)²

m² − n² = (tan A + sin A + tan A − sin A) (tan A + sin A − tan A + sin A)

m² − n²  = (2 tan A)(2 sin A)

m² − n²  = 4 tan A sin A    ...(1)

Also,

`4 sqrt (mn) = 4 sqrt ((tan A + sin A)( tan A - sin A))`

= `4 sqrt(tan^2 A - sin^2 A)`

= `4 sqrt((sin^2 A)/(cos^2 A) - sin^2 A)`

= `4 sin A sqrt((1 - cos^2 A)/(cos^2 A))`

= `4 sin A sqrt((sin^2 A)/(cos^2 A))`

= `4 sin A * (sin A)/(cos A)`

= `4 sin A . tan A`    ...(2)

Using equation (1) and equation (2) we get the required conditions.
Hence proved.