Question

Solve the problem : show that m²-n²=4 rootmn

Answer 1

here is the answer....

Answer 2

Heya,

Let theta be 'A'

Given=> tanA + sinA = m. .... equation 1
tanA - sinA = n. ...... equation 2

Squaring both the equation, we get;
tan²A + sin²A + 2tanAsinA = m². .... equation 3
tan²A + sin²A - 2tanAsinA = n². ...... equation 4

subtracting equation 4 minus equation 3

(tan²A + sin²A + 2tanAsinA) - (tan²A + sin²A - 2 tanAsinA) = m² - n²
=> tan²A + sin²A + 2tanAsinA - tan²A - sin²A + 2 tanAsinA
=> 4tanAsinA = m² - n²

Squaring both side...

=> 16tan²Asin²A = (m² - n²)²
=> 16 tan²A(1-cos²A) = (m² - n²)²
=> 16 tan²A - tan²Acos²A = (m² - n²)²
=> 16 tan²A - sin²A/cos²Ax cos2A = (m² - n²)²
=> 16 tan²A - sin²A = (m² - n²)²
=> 16 (tanA +sinA) (tanA - sinA) = (m² - n²)²
=> 16mn = (m² - n²)²
=> m² - n² = 4√mn

Hence Proved

Hope this helps....:)