Equation of a line passing through points (cosθ, sin θ) and (cos Φ, sin Φ). My solution to the above question gives slope of the line(m) as:
m= (sinΦ - sinθ)/(cosΦ - cosθ) = -cot(Φ/2 + θ/2)

And the equation thereupon is:
y = -cot(Φ/2 + θ/2).x + c

The value of c comes out to be:
c= -cot(Φ/2 + θ/2).cos Φ - sin Φ

 (on putting in the co-ordinates of second point)

Is the expression for 'c' correct? I have some doubts regarding that.
Post-Script:
There were no 'Straight-line' tags
 A: For such a question, I think it's worth having a geometrical view.
Let $\alpha:=(\theta+\Phi)/2$.
Take a look at the following picture :

Triangle $AOB$ being isosceles, vector $\binom{\cos \alpha}{\sin \alpha}$ (in red) bissects angle $AOB$ therefore is orthogonal to vector $\vec{AM}=\binom{x- \cos \Phi}{y-\sin \Phi}$ where $M(x,y)$ is any point of straight line $AB$. This orthogonality means that the dot product of these two vectors is zero. As a consequence, the equation of straight line $AB$ is :
$$(x- \cos \Phi)\cos \alpha+(y-\sin \Phi)\sin \alpha=0$$
It remains to expand this equation into :
$$x \cos \alpha+y \sin \alpha=\underbrace{\cos \Phi\cos \alpha+ \sin \Phi\sin \alpha}_{\cos(\Phi-\alpha)}$$
i.e.,

$$x \cos \alpha+y \sin \alpha=\cos \beta$$
  where $\beta:=\Phi-\alpha=(\Phi-\theta)/2$ (half angle AOB).

A: Without checking the trig identities this looks like a mostly correct approach.
$$y=mx+c \rightarrow m=\frac{\Delta y}{\Delta x}$$
For finding $c$ few details are provided. I would normally plug in one of the points and solve for $c$. It appears that you may have attempted this?
$$ \sin\Phi = m\cos \Phi+c$$
$$ c =   \sin \Phi - m \cos\Phi$$
seems you may have a sign issue with your choice of $c$
A: Use the line formula below
$$(x_2-x_1)y-(y_2-y_1)x = x_2y_1 - x_1y_2$$
for the two given points (cosθ, sin θ) and (cos Φ, sin Φ),
$$(\cos Φ -\cos θ)y- (\sin Φ -\sin θ) x  = \cos Φ \sin θ -\cosθ \sin Φ = -\sin(Φ -θ)$$
Use the trigonometric identities 
$$\cos Φ -\cos θ=-2\sin \frac{ Φ+ θ}{2} \sin \frac{ Φ- θ}{2}$$
$$\sin Φ -\sin θ=2\cos \frac{ Φ+ θ}{2}\sin \frac{ Φ- θ}{2} $$
$$\sin(Φ -θ) = 2 \cos \frac{ Φ- θ}{2}\sin \frac{ Φ- θ}{2}$$
to eliminate the common factor $2\sin \frac{ Φ- θ}{2}$ and express the line equation as,
$$\sin \frac{ Φ+ θ}{2}y
+ \cos \frac{ Φ+ θ}{2}x=\cos \frac{ Φ- θ}{2}$$
A: A sketch shows that we need half-sum and half-difference of angles $ (Φ ,\theta ) $ HS and HD (bisector/its pedal perpendicular) respectively in polar-normal form of a straight line which is the correct choice needed here. So we have 
$$ x \cos HS + y \sin HS = \cos HD $$
