Finding equation of chord of hyperbola. Equation of chord of hyperbola joining points $(a\sec\phi,b\tan\phi)$ and $(a\sec\phi_1,b\tan\phi_1) $  $$y-b\tan\phi=\frac{b\tan\phi-b\tan\phi_1}{a\sec\phi-asec \phi_1}(x-a\sec\phi) $$ This reduces to $$\frac{y}{b}\sin\Big(\frac{\phi+\phi_1}{2}\Big)-\frac{x}{a}\cos\Big(\frac{\phi-\phi_1}{2}\Big)=\tan\phi \sin\Big(\frac{\phi+\phi_1}{2}\Big)-\sec\phi \cos\Big(\frac{\phi-\phi_1}{2}\Big) $$
Now I want to reduce this in the form$$-\frac{y}{b}\sin\Big(\frac{\phi+\phi_1}{2}\Big)+\frac{x}{a}\cos\Big(\frac{\phi-\phi_1}{2}\Big)=\cos\Big(\frac{\phi+\phi_1}{2}\Big) $$
How to get this form ? Most of the books use this one but they dont give the proof which I seek.
 A: If you multiply both sides of $$\frac{y}{b}\sin\Big(\frac{\phi+\phi_1}{2}\Big)-\frac{x}{a}\cos\Big(\frac{\phi-\phi_1}{2}\Big)=\tan\phi \sin\Big(\frac{\phi+\phi_1}{2}\Big)-\sec\phi \cos\Big(\frac{\phi-\phi_1}{2}\Big)$$ by $-1$ you get:
$$\frac{x}{a}\cos\Big(\frac{\phi-\phi_1}{2}\Big)-\frac{y}{b}\sin\Big(\frac{\phi+\phi_1}{2}\Big)=\sec\phi \cos\Big(\frac{\phi-\phi_1}{2}\Big)-\tan\phi \sin\Big(\frac{\phi+\phi_1}{2}\Big)$$
which means that all we need to show is that:
$$ \sec\phi \cos\Big(\frac{\phi-\phi_1}{2}\Big)-\tan\phi \sin\Big(\frac{\phi+\phi_1}{2}\Big)=\cos\left(\frac{\phi+\phi_1}{2}\right) $$
which you can prove by doing the following:$$
\sec \phi\left[ \cos\Big(\frac{\phi-\phi_1}{2}\Big)-\sin\phi \sin\Big(\frac{\phi+\phi_1}{2}\Big)\right]$$
then decompose the cosine and sine using the addition formula:
$$ \sec \phi \left[
\cos\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi_1}{2}\right)+\sin\left(\frac{\phi}{2}\right)\sin\left(\frac{\phi_1}{2}\right)
\\
-\sin\phi\left( \sin\left(\frac{\phi}{2}\right) \cos\left(\frac{\phi_1}{2}\right)+\cos\left(\frac{\phi}{2}\right)\sin\left(\frac{\phi_2}{2}\right)
\right)
\right] $$
then use the identity $\sin\phi=2\sin\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)$ and collect some like terms together to get:
$$
 \sec \phi \left[
\cos\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi_1}{2}\right)\left[ 1-2\sin^2\left(\frac{\phi}{2}\right)\right]+\sin\left(\frac{\phi}{2}\right)\sin\left(\frac{\phi_1}{2}\right)\left[ 1-2\cos^2\left(\frac{\phi}{2}\right) \right]
\right]
$$
then you can use the identities
$$\cos\phi=2\cos^2\left(\frac{\phi}{2}\right)-1=1-2\sin^2\left(\frac{\phi}{2}\right)$$
to get:
$$ \sec \phi \left[
\cos\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi_1}{2}\right)\left[\cos\phi\right]+\sin\left(\frac{\phi}{2}\right)\sin\left(\frac{\phi_1}{2}\right)\left[-\cos\phi \right]
\right]$$
which gives:
$$
\cos\left(\frac{\phi}{2}\right)\cos\left(\frac{\phi_1}{2}\right)-\sin\left(\frac{\phi}{2}\right)\sin\left(\frac{\phi_1}{2}\right)
$$
which, of course is:
$$ \cos\left(\frac{\phi+\phi_1}{2}\right) $$
which is what we needed to show.
A: The rhs of your second and third equations should be equal and opposite, as one can verify:
$$
\tan\phi \sin { \phi + \phi_1 \over 2} - \sec \phi \cos {\phi - \phi_1 \over 2} + \cos {\phi + \phi_1 \over 2} =
{1 \over \cos \phi} \left( \sin \phi \sin {\phi+ \phi_1 \over 2} - \cos { \phi- \phi_1 \over 2} + \cos \phi \cos { \phi+ \phi_1 \over 2}
\right)=
{1 \over \cos \phi} \left( \cos {  \phi - \phi_1 \over 2} - \cos {\phi - \phi_1 \over 2} \right) = 0
$$
where in the third line, I used the trig identity
$$
\cos (a - b) = \cos a \cos b + \sin a \sin b
$$
A: Another method which does not require ingenious use of trigonometric identities goes as follows
Equation of tangent to hyperbola at point $(asec \ A,btan \ A)$ is $$\frac{x}{a}sec \ A-\frac{y}{b}tan\ A=1 $$ 
Equation of tangent to hyperbola at point $(asec \ B,btan \ B)$ is $$\frac{x}{a}sec \ B-\frac{y}{b}tan\ B=1 $$ 
The intersection of these two tangents is the point $$\Bigg(a\frac{cos\frac{A-B}{2}}{cos\frac{A+B}{2}},b\frac{sin\frac{A+B}{2}}{cos\frac{A+B}{2}}\Bigg) $$ 
The equation of chord of contact from a point on a conic is $T=0$. Hence equation of chord is $$\frac{x}{a^2}a\frac{cos\frac{A-B}{2}}{cos\frac{A+B}{2}} -\frac{y}{b^2}b\frac{sin\frac{A+B}{2}}{cos\frac{A+B}{2}}=1 $$ Which on simplification gives $$\frac{x}{a}{cos\frac{A-B}{2}}-\frac{y}{b}sin\frac{A+B}{2}=cos\frac{A-B}{2} $$
