Question

if cot theta(1+sin theta)=4m and cot theta(1- sin theta)=4n prove that (m^2-n^2)^2=mn

Answer 1

we have
m=cot ϴ(1+sinϴ)/4
n=cot ϴ(1-sinϴ)/4
to prove

(m²-n²)²=mn
SO 
first of all we should simplify the RHS

mn=[cot ϴ(1+sinϴ)/4][cot ϴ(1-sinϴ)/4]

mn=cot² ϴ(1-sin²ϴ)/16                {cot² ϴ=cos²ϴ/sin²ϴ}

mn=cos²ϴ/sin²ϴ*(1-sin²ϴ)/16

mn=cos²ϴ*(1-sin²ϴ)/16sin²ϴ

mn=cos²ϴ*(cos²ϴ)/16sin²ϴ             { 1-sin²ϴ=cos²ϴ}

mn=cos↑4ϴ/16sin²ϴ 

NOW LHS

(m²-n²)²

[cot² ϴ(1+sin²ϴ)/16- cot ²ϴ(1-sin²ϴ)/16]²

{[cot² ϴ(1+sin²ϴ) - cot ²ϴ(1-sin²ϴ)]/16}²

[(4sinϴcot² ϴ)/16]²

cos↑4ϴ/16

hence LHS = RHS

Answer 2

we have
m=cot ϴ(1+sinϴ)/4
n=cot ϴ(1-sinϴ)/4
to prove
(m²-n²)²=mn
SO
first of all we should simplify the RHS
mn=[cot ϴ(1+sinϴ)/4][cot ϴ(1-sinϴ)/4]
mn=cot² ϴ(1-sin²ϴ)/16 {cot² ϴ=cos²ϴ/sin²ϴ}
mn=cos²ϴ/sin²ϴ*(1-sin²ϴ)/16
mn=cos²ϴ*(1-sin²ϴ)/16sin²ϴ
mn=cos²ϴ*(cos²ϴ)/16sin²ϴ { 1-sin²ϴ=cos²ϴ}
mn=cos↑4ϴ/16sin²ϴ
NOW LHS
(m²-n²)²
[cot² ϴ(1+sin²ϴ)/16- cot ²ϴ(1-sin²ϴ)/16]²
{[cot² ϴ(1+sin²ϴ) - cot ²ϴ(1-sin²ϴ)]/16}²
[(4sinϴcot² ϴ)/16]²
cos↑4ϴ/16
hence LHS = RHS