The tangent to the ellipse and circle

Let P(a cos θ , b sinθ ) be a point on the ellipse (x²/a²) + (y²/b²) = 1 .
If the tangent at P to the ellipse touches the circle x² + y² = c²
Where 0 < b < c < a , show that tan² θ = [b²(c²-a²)/a²(b²-c²)]
Solution
The tangent equation of ellipse at P is
b(cosθ)x+a(sinθ)y=ab...(1)
y=[ab-b(cosθ)x]/asinθ
substitute into the circle x² + y² = c²
x² + {[ab-b(cosθ)x]/asinθ}² = c²
(asinθx)² + [ab-b(cosθ)x]² = (asinθc)²
(a^2sin^θ+b^2cos^2θ)x^2-2ab^2cosθx+a^2b^2-a^2c^2sin^2θ=0
Since there is only one root of this equation
Discriminent=0
4a^2b^4cos^2θ-4(a^2sin^θ+b^2cos^2θ)(a^2b^2-a^2c^2sin^2θ)=0
b^4cos^2θ-(a^2sin^θ+b^2cos^2θ)(b^2-c^2sin^2θ)=0
Expand
b^4cos^2θ-(a^2b^2sin^θ-a^2c^2sin^4θ+b^4cos^2θ-b^2c^2sin^2θcos^2θ)=0
-(a^2b^2sin^θ-a^2c^2sin^4θ-b^2c^2sin^2θcos^2θ)=0
a^2b^2sin^θ=a^2c^2sin^4θ+b^2c^2sin^2θcos^2θ
a^2b^2=a^2c^2sin^2θ+b^2c^2θcos^2θ
a^2b^2sec^2θ=a^2c^2tan^2θ+b^2c^2θ
a^2b^2(1+tan^2θ)=a^2c^2tan^2θ+b^2c^2θ
(a^2b^2-a^2c^2)tan^2θ=b^2c^2-a^2b^2
a^2(b^2-c^2)tan^2θ=b^2(c^2-a^2)
tan² θ = [b²(c²-a²)/a²(b²-c²)]