Question

If cot θ(1 + sin θ) = 4m and cot θ(1 – sin θ) = 4n then prove that (m² – n²)² = m

Answer 1

Given `cot theta(1 + sin theta)` = 4 m

`cot theta(1 - sin theta)` = 4 n

(4m)² – (4n)² = `cot^2theta(1 + sintheta)^2 - cot^2theta(1 - sintheta)^2`

16m² – 16n² = `cot^2theta[(1 + sintheta)^2 - (1 - sin theta)^2]` 

16(m² – n²) = `cot2theta [(1 + 2sintheta + sin^2theta) - (1 - 2sintheta + sin^2theta)]` 

16(m² – n²) = `cot^2theta [1 + 2sintheta + sin^2theta - 1 + 2sintheta - sin^2theta]`

16(m² – n²) = `cot^2theta*4sintheta`

4(m² – n²) = `cot^2theta sintheta`

Squaring on both sides

16(m² – n²)² = `cot^4theta sin^2theta`

16(m² – n²)² = `cot^2theta*cot^2theta sin^2theta`

16(m² – n²)² = `cot^2theta("cosec"^2theta - 1) sin^2theta`

16(m² – n²)² = `cot^2theta ("cosec"^2theta sin^2theta - sin^2theta)`

16(m² – n²)² = `cot^2theta(1 - sin^2theta)`

16(m² – n²)² = `cot^2theta cos^2theta`  ......(1)

4m . 4n = `cot theta(1 + sin theta) cottheta(1 - sin theta)`

16 mn = `cot^2theta(1 - sin^2theta)`

16 mn = `cot^2theta cos^2theta`  ......(2)

Equations (1) and (2), we have

16(m² – n²)² = 16 mn

(m² – n²)² = mn