Question

If tan A + sin A = m and tan A - sin A = n, then show that m² - n² = 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) = 4sqrt((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 sqrt((sin A)/(cos A))`

= 4 sin A. tan A         ....(2)

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