if (asquare-bsquare)sintheta=2abcostheta=asquare+bsquare then prove that tantheta=asquare-bsquare/2ab
Is that supposed to say:
(a^2 -b^2)sinØ = 2ab(cosØ) =a^2 + b^2
then prove : tanØ = (a^2 - b^2)/(2ab) ?
then using:
(a^2 - b^2)sinØ = 2abcosØ
sinØ/cosØ = 2ab/(a^2 - b^2)
tanØ = 2ab/(a^2-b^2)
using:
(a^2 -b^2)sinØ = a^2 + b^2
sinØ = (a^2 + b^2)/(a^2-b^2)
using:
2ab(cosØ) =a^2 + b^2
cosØ = (a^2+b^2)/(2ab)
then sinØ/cosØ = [(a^2 + b^2)/(a^2-b^2)]/[ (a^2+b^2)/(2ab)
tanØ = 2ab/(a^2-b^2) just as before
then cotØ 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 Ø , invariably gives you
tanØ = 1
so Ø = 45° or π/4