Wednesday
June 19, 2013

# Homework Help: math

Posted by akhtar on Saturday, August 18, 2012 at 8:41am.

if (asquare-bsquare)sintheta=2abcostheta=asquare+bsquare then prove that tantheta=asquare-bsquare/2ab

• math - Reiny, Saturday, August 18, 2012 at 9:38am

Is that supposed to say:
(a^2 -b^2)sinŲ = 2ab(cosŲ) =a^2 + b^2
then prove : tanŲ = (a^2 - b^2)/(2ab) ?

then using:
(a^2 - b^2)sinŲ = 2abcosŲ
sinŲ/cosŲ = 2ab/(a^2 - b^2)
tanŲ = 2ab/(a^2-b^2)

using:
(a^2 -b^2)sinŲ = a^2 + b^2
sinŲ = (a^2 + b^2)/(a^2-b^2)

using:
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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