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.

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 $$\frac{x}{a}{cos\frac{A-B}{2}}-\frac{y}{b}sin\frac{A+B}{2}=cos\frac{A-B}{2} $$
