Trouble with tedious algebra (Oxford 1992 Admissions Test 2 1992) 
(i) Show that the condition that the points $P$ $(a\cos A,b\sin A)$ and $Q$ $(a\cos B,b\sin  B )$
should subtend a right angle at O is
$$a^2\cos A\cos B+b^2\sin A\sin  B=0$$
(ii) Let S be a circle centre $O$ and radius $C$. Find the equation of the tangent to S at
the point $(C\cos  t, C\sin t)$.
(iii) If $C = \dfrac{ab}{\sqrt{a^2+b^2}}$, show that the points where a tangent to S cuts the ellipse $$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
subtend a right angle at $O$.

Part (i) can be done by considering gradients of lines from $O$ to each of the points, multiplying them and setting equal to $-1$
Part (ii) gives the result $x\cos t+y\sin t=C$
With part (iii), I have attempted to simply rearrange for $y$ in the tangent equation, subbing into the ellipse equation and trying to solve for $x$ but this results in a huge amount of tedious algebra and I am unable to simplify it properly.
I then attempted to parameterize the ellipse in the form $x=a\cos T, y=b\sin T$ but I am unsure how to go from there to solve for the points. I did attempt to use harmonic addition and this does give some exact solutions in terms of $a,b,t$ but to simplify requires identities for $\sin(\arccos(x))$ and other similar identities like $\arctan\left(\frac{b}{a}\tan t\right)$.
Some useful information may be that we can let the points of intersections be $R(a \cos P,b\sin P)$ and $L(a\cos Q,b\sin Q)$ and substitute these points into the equation for the tangent. Now my problem is taking those equations and getting the equation we want from (i)

Another way: I might say this way is "more obvious" in the sense that it is easy to think of. You "just" sit down and crunch through the algebra as if it were an ordinary simultaneous equation job. Though, one must be very careful with the algebra.
The cases where $\cos(t)=0$ or $\sin(t)=0$ can be very easily handled. Suppose now that neither is zero. Then we arrange the tangent line as: $$y=C\csc(t)-x\cot(t)$$And the ellipse equation can be arranged as: $$b^2x^2+a^2y^2-a^2b^2=0$$Let $y=C\csc(t)-x\cot(t)$ in the ellipse equation: $$\begin{align}b^2x^2+a^2[C^2\csc^2(t)+x^2\cos^2(t)\csc^2(t)-2C\cos(t)\csc^2(t)\cdot x]-a^2b^2&=0\\x^2[b^2\sin^2(t)+a^2\cos^2(t)]+a^2C^2-a^2b^2\sin^2(t)-2a^2C\cos(t)\cdot x&=0\\\frac{x^2}{a^2}[b^2\sin^2(t)+a^2\cos^2(t)]+\frac{a^2b^2}{a^2+b^2}-b^2\sin^2(t)-\frac{2ab\cos(t)}{\sqrt{a^2+b^2}}\cdot x&=0\end{align}$$Now let $x=a\cos\vartheta$ for some $\vartheta$ to be determined: $$\cos^2\vartheta[b^2\sin^2(t)+a^2\cos^2(t)]+b^2\cdot\frac{a^2\cos^2(t)-b^2\sin^2(t)}{a^2+b^2}-\frac{2a^2b\cos(t)}{\sqrt{a^2+b^2}}\cos\vartheta=0$$We use the quadratic formula. The "$b^2-4ac$" term involves a very nice difference-of-two-squares: $$\begin{align}\frac{4b^2}{a^2+b^2}a^4\cos^2(t)-&\frac{4b^2}{a^2+b^2}[b^2\sin^2(t)+a^2\cos^2(t)][a^2\cos^2(t)-b^2\sin^2(t)]\\&=\frac{4b^2}{a^2+b^2}[a^4\cos^2(t)(1-\cos^2(t))+b^4\sin^4(t)]\\&=\frac{4b^2}{a^2+b^2}\sin^2(t)\cdot[a^4\cos^2(t)+b^4\sin^2(t)]\end{align}$$The quadratic formula then yields: $$\begin{align}\cos\vartheta&=\frac{\frac{2a^2b\cos(t)}{\sqrt{a^2+b^2}}\pm\frac{2b\sin(t)}{\sqrt{a^2+b^2}}\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{2[b^2\sin^2(t)+a^2\cos^2(t)]}\\&=\frac{b}{\sqrt{a^2+b^2}}\cdot\frac{a^2\cos(t)\pm\sin(t)\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{b^2\sin^2(t)+a^2\cos^2(t)}\end{align}$$A near-identical computation, arising from letting $x=C\sec(t)-y\tan(t)$, cancelling the $\sec^2$ terms, and dividing by $b^2$ to get a quadratic formula in $\sin\vartheta$, then solving, gives: $$\sin\vartheta=\frac{a}{\sqrt{a^2+b^2}}\cdot\frac{b^2\sin(t)\pm\cos(t)\sqrt{a^4\cos^2(t)+b^4\sin^2(t)}}{b^4\sin^2(t)+a^4\cos^2(t)}$$Now we use the hint of part $(i)$ and the difference of two squares: $$\begin{align}a^2\cos(\vartheta_+)\cos(\vartheta_-)+b^2\sin(\vartheta_+)\sin(\vartheta_-)&=\underset{K}{\underbrace{\left(\frac{ab}{[b^4\sin^2(t)+a^4\cos^2(t)]\sqrt{a^2+b^2}}\right)^2}}\\&\times\Big[a^4\cos^2(t)-\sin^2(t)[b^4\sin^2(t)+a^4\cos^2(t)]\\&+b^4\sin^2(t)-\cos^2(t)[b^4\sin^2(t)+a^4\cos^2(t)]\Big]\\&=K\cdot\Big[a^4\cos^2(t)(1-\sin^2(t))+b^4\sin^2(t)(1-\cos^2(t))\\&-a^4\cos^4(t)-b^4\sin^4(t)\Big]\\&=0\end{align}$$
So the intersection points involving $a\cos\vartheta_{\pm},b\sin\vartheta_{\pm}$ (the pairing of which will vary in general) subtend a right angle, as desired.
