Obtain tangent line of $P(a\cos\phi, b\sin \phi)$ on $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ 1) Obtain the equation of the tangent $P(a\cos\phi, b\sin \phi)$ on the ellipse  $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
2) If the tangent at P meets the axes at $TT^\prime$ and the diameter through P meets the ellipse again at $P^\prime$. Show that $$\tan TP^\prime T^\prime=\frac{2OT\cdot OT^\prime}{a^2+b^2+OP^2}$$
My attempt:
Equation of tangent to an ellipse: $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1\quad\\\frac{x\cos\phi}{a}+\frac{y\sin\phi}{b}=1 \quad \color{red}{P(a\cos\phi, b\sin\phi) }$$
I have a serious challenge with the second part of the question I made the following deductions from the diagram:
${TT^\prime}^2={OT'}^2+{OT}^2\\PP'=OP+OP'\\$
I also considered finding the angles between lines $P'T $ and $P'T'$ but don't seem to be making any headway. 
Any hint on how this can be done?
 A: 
The tangent line of the point $P(a\cos\phi,b\sin\phi)$ is given by
$$\frac{x\cos\phi}{a}+\frac{y\sin\phi}{b} =1$$
from which to get $T(\frac a{\cos\phi},0)$ and $T'(0, \frac b{\sin \phi})$, along with
$P'(-a\cos\phi,-b\sin\phi)$. Then, the tangents of the lines $P'T$, $P'T'$, $P'P$ are, respectively,
$$m_{P'T} = \frac{b\sin\phi\cos\phi}{a(1+\cos^2\phi)},\>\>\> 
m_{P'T'} = \frac{b(1+\sin^2\phi)}{a\sin\phi\cos\phi},\>\>\>m_{P'P} = \frac{b\sin\phi}{a\cos\phi}$$
Then, evaluate
$$\tan PP'T = \frac{ m_{P'P}  - m_{P'T}}{1+m_{P'P}\cdot m_{P'T}}= \frac{OT\cdot OT^\prime\sin^2\phi}{a^2+ OP^2 } $$
$$\tan T'P'P = \frac{ m_{P'T'}  - m_{P'P}}{1+m_{P'T'}\cdot m_{P'P}}= \frac{OT\cdot OT^\prime\cos^2\phi}{b^2+ OP^2 }$$
Finally
$$\tan T'P'T = \frac{ \tan PP'T + \tan T'P'P }{1- \tan PP'T \cdot \tan T'P'P} 
=\frac{2OT\cdot OT^\prime}{a^2+b^2+OP^2}$$
A: $\frac {x}{a\sec\phi} + \frac {y}{b\csc\phi} = 1$ is the equation of a line in intercept - intercept form.
$T = (a\sec \phi,0), T' = (0,b\csc \phi)$
$P'$ is on the opposite side of the elipse from $P.\  P' = (-a\cos\phi, -b\sin\phi)$
$\|(T-P')\times(T'-P')\| = \|T-P'\|\|T'-P'\|\sin \theta\\
(T-P')\cdot(T'-P') = \|T-P'\|\|T'-P'\|\cos \theta$
Have you learned about dot products and cross products?
$\tan \theta = \frac {(a\sec\phi + a\cos\phi)(b\csc\phi + b\sin\phi) - (a\cos\phi)(b\sin\phi)}{(a\sec\phi + a\cos\phi)(a\cos\phi) + (b\sin\phi)(b\csc\phi+ b\sin\phi)}\\
\frac{ab(\sec\phi\csc\phi + \tan\phi + \cot\phi)}{a^2(1+\cos^2\phi) + b^2(1+\sin^2\phi)}$
$\csc\phi\sec\phi = \cot\phi + \tan\phi = OT\cdot OT'$
And $a^2\cos^2\phi + b^2\sin^2\phi = OP^2$
