Question

If (a2-b2)sinθ+2abcosθ = a2+b2 find the value of tanθ in terms of ‘a’ and ‘b’.

Answer 1

$(a^2-b^2)Sin\theta+2abCos\theta=a^2+b^2\\\\Let\ a=kCos\phi,\ \ b=kSin\phi,\ \ a^2+b^2=k^2,\ \ tan\phi=b/a\\\\k^2(Cos^2\phi-Sin^2\phi)Sin\theta+2k^2Cos\phi\ Sin\phiSin\theta=k^2\\\\Cos2\phi\ Sin\phi+Sin2\phi\ Cos\theta=1\\\\Sin(2\phi+\theta)=1\\\\ \theta=\pi/2-2\phi\\\\tan\ \theta=cot\ 2\phi\\\\=\frac{1+tan^2\phi}{2tan\phi}=\frac{a^2+b^2}{2ab}$

Answer 2

Divide both sides by cos(A) to get:

(a^2 - b^2)tan(A) + 2ab = (a^2 + b^2)sec(A)

Square both sides:

(a^2 - b^2)^2 * tan^2(A) + 4ab(a^2 - b^2)tan(A) + 4a^2b^2 = (a^2 + b^2)^2 * sec^2(A)

Since 1 + tan^2(A) = sec^2(A):

(a^2 - b^2)^2 * tan^2(A) + 4ab(a^2 - b^2)*tan(A) + 4a^2b^2 = (a^2 + b^2)^2 * (1 + tan^2(A))

((a^2 - b^2)^2 - (a^2 + b^2)^2)tan^2(A) + 4ab(a^2 - b^2)*tan(A) + 4a^2b^2 - (a^2 + b^2)^2 = 0

(-4a^2b^2)tan^2(A) + 4ab(a^2 - b^2)tan(A) + 2a^2b^2 - a^4 - b^4 = 0

tan^2(A) + (b^2 - a^2)/(ab) * tan(A) - 1/2 + (a^2)/(4b^2) + (b^2)/(4a^2) = 0

Let u = tan(A)

u = [(a^2 - b^2)/(ab) +/- sqrt((b^2 - a^2)^2/(a^2b^2) - 4((a^2)/(4b^2) + (b^2)/(4a^2) - 1/2))]/2

u = ((a^2 - b^2)/(ab) +/- sqrt((b^2/(a^2) - 2 + (a^2)/(b^2) - (a^2)/(b^2) - (b^2)/(a^2) + 2)/2

u = tan(A) = (a/b - b/a +/- sqrt(0))/2

tan(A) = (a/b - b/a)/2 = (a^2 - b^2)/(2ab)