Question

If (a²- b²)sin theta + 2ab cos theta = a²+ b², prove that tan theta = a²- b² / 2ab

Answer 1

There can be many methods for this:

one is:

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)