Question

if tana+sina=m &tana-sina=n then proved that m^2 - n^2 =4√mn​

Answer 1

Answer:

answer for the given problem is given

Answer 2

Answer:

Step-by-step explanation:

tan a + sin a = m

tan a - sin a = n

m^2 - n^2 = (tan a + sin a)^2 - (tan a - sin a)^2

                 = tan^2 a + sin^2 a + 2 tan a sin a - (tan^2 a + sin^2 a - 2 tan a sin a)

                 = tan^2 a + sin^2 a + 2 tan a sin a - tan^2 a - sin^2 a + 2 tan a sin a

                 = 4 tan a sin a                                                - (1)

4√mn = 4√(tan a + sin a)(tan a - sin a)

          = 4√(tan^2 a - sin^2 a)

          = 4√(sin^2 a/cos^2 a - sin^2 a)

          = 4√sin^2 a( 1/cos^2 a - 1)

          = 4 sin a√(sec^2 a - 1)

          = 4 sin a√(tan^2 a)

          = 4 sin a tan a

          = 4 tan a sin a                                                      - (2)

From (1) and (2) -

m^2 - n^2 = 4√mn