Question

If tanθ+ sinθ= m and tanθ- sinÆŸ = n,show that m 2 - n 2 = 4√mn

Answer 1

m= tanθ + sinθ ; n = tanθ – sinθ
Squarring on both sides for the above equations.
m² = (tanθ + sinθ)²------equation(1)
n²=(tanθ – sinθ)² --------equation(2)
Now Substract both the equations (1) and (2)

m² – n² = (tanθ + sinθ)² – (tanθ – sinθ)²
m² – n² = tan²θ + sin²θ + 2sinθ tanθ – tan²θ – sin²θ + 2sinθ tanθ
cancel (+ tan²θ,-tan²θ and +sin²θ, -sin²θ), we get,
m² – n²=4sinθtanθ ---eq(a)

Now multiply (m) and (n)

mn = (tanθ + sinθ)(tanθ – sinθ)
mn=tan²θ – sin²θ
mn=sin²θ/cos²θ – sin²θ
Take sin²θ as common
mn=sin²θ(1/cos²θ – 1)
mn=sin²θ(1–cos²θ/cos²θ)
mn=sin²θ(sin²θ/cos²θ)
mn= sin4θ/cos²θ

4√mn= 4 root(sin4θ/cos²θ)
=4 sin²θ/cosθ
=4sinθtanθ------(b)

Therefore from equations (a) and (b), we get

m² – n² = 4√mn