Question

Cotα(1_sinα)=4m or cotα(1-sinα)=4n then Prove(square m-square n) whole square=mn

Answer 1

Answer: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

hope it helps u

Step-by-step explanation: