Question

If tanA+sinA=m , tanA-sinA=n, prove that (m^2-n^2)^2=16

Answer 1

Given: tanФ+sinФ=m and tanФ-sinФ=n

m² = tan²Ф + sin²Ф + 2 tanФ sinФ
n² = tan²Ф + sin²Ф - 2 tanФ sinФ


Subtracting m² with n²,
m² - n² = 4tanФ sinФ------(i)


Finding mn,
mn = tan²Ф - sin²Ф
= sin²Ф/cos²Ф - sin²Ф/1
= sin²Ф(1 - cos²Ф) / cos²Ф
= (sin²Ф / cos²Ф) x sin²Ф
= tan²Ф sin²Ф-------(ii)


Now ATQ,
(m²-n²)² = (4tanФ sinФ)²     [from (i) and (ii)]
= 16tan²Ф sin²Ф
= 16mn
(HENCE PROVED)