Question

if costheta (1+sintheta)=4m and costheta (1-sintheta)=4n then prove that (msquare-nsquare) square=mm​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

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

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