Posted by akhtar on Saturday, August 18, 2012 at 8:41am.
Is that supposed to say:
(a^2 -b^2)sinŲ = 2ab(cosŲ) =a^2 + b^2
then prove : tanŲ = (a^2 - b^2)/(2ab) ?
(a^2 - b^2)sinŲ = 2abcosŲ
sinŲ/cosŲ = 2ab/(a^2 - b^2)
tanŲ = 2ab/(a^2-b^2)
(a^2 -b^2)sinŲ = a^2 + b^2
sinŲ = (a^2 + b^2)/(a^2-b^2)
2ab(cosŲ) =a^2 + b^2
cosŲ = (a^2+b^2)/(2ab)
then sinŲ/cosŲ = [(a^2 + b^2)/(a^2-b^2)]/[ (a^2+b^2)/(2ab)
tanŲ = 2ab/(a^2-b^2) just as before
then cotŲ would be (a^2-b^2)/(2ab) not the tangent.
To be true......
The only fraction equal to its reciprocal is 1
so a^2 - b^2 = 2ab
a^2 - 2ab - b^2 = 0
solving for a using the quadratic formula and simpliflying , I got
a = b ± √2 b
check: picking any value of b, finding a, and then evaluating
tan Ų , invariably gives you
tanŲ = 1
so Ų = 45° or π/4
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