(a^2 - b^2) sin theta+ 2abcos thetha= a^2 + b^2 then find tan theta

Answer 1

(a^2-b^2)sinθ + 2abcosθ = (a^2+b^2)
(a^2-b^2)tanθ + 2ab = (a^2+b^2)secθ
(a^2-b^2)^2 tan^2θ = ((a^2+b^2)secθ-2ab)^2
(a^2-b^2)^2(sec^2θ-1) = (a^2+b^2)^2sec^2θ - 4ab(a^2+b^2)secθ + 4a^2b^2
((a^2-b^2)^2-(a^2+b^2)^2)sec^2θ + 4ab(a^2+b^2)secθ + 4a^2b^2 = 0
-4a^2b^2sec^2θ + 4ab(a^2+b^2)secθ + 4a^2b^2 = 0
ab sec^2θ - 4(a^2+b^2)secθ - ab = 0

secθ = (2(a^2+b^2)±√(4(a^2+b^2)^2+a^2b^2))/ab
so, tan^2θ = sec^2θ-1

I can't help feeling there's a simpler, more elegant way...

Answer 2

or, how about this?

(a^2-b^2)sinθ + 2abcosθ = (a^2+b^2)
sinθ + 2ab/(a^2-b^2)cosθ = (a^2+b^2)/(a^2-b^2)

Multiply the the fractions top and bottom by 1/a^2, and we have
sinθ + 2(b/a)/(1-(b/a)^2)cosθ = (1+(b/a)^2)/(1-(b/a)^2)

Now let tanx = b/a
sinθ + tan(2x)cosθ = (1+tan^2x)/(1-tan^2x)

now multiply by cos(2x)=cos^2x-sin^2x:
cos2x sinθ + sin2x cosθ = (cos^2x-sin^2x) * sec^2x/(1-tan^2x)
sin(2x+θ) = (1-tan^2x)/(1-tan^2x)
sin(2x+θ) = 1
2x+θ = π/2
θ = π/2-2x

tanθ = cot(2x)
= (1-(b/a)^2)/(2(b/a))
= (a^2-b^2)/(2ab)