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IF cotA(1+sinA) = 4m And Cora(1-sinA)= 4nThen prove. (m^2-n^2) = mn

Amartya S. Das , 8 Years ago
Grade 11
anser 1 Answers
Tyler Belvins
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
 
Last Activity: 7 Years ago
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