Showing $c^2+a^2= \csc^2\theta$, given $a^2+b^2+2ab \cos\theta=1$, $c^2+d^2+2cd \cos\theta=1$, $(ac+bd)+(ad+bc)\cos\theta=0$

Question

Given that $$\begin{align} a^2+b^2+2ab \cos\theta &=1 \\ c^2+d^2+2cd \cos\theta &=1 \\ (ac+bd)+(ad+bc)\cos\theta &=0 \end{align}$$ Prove that $c^2+a^2 = \csc^2\theta$

I tried to solve this question by separating $\cos\theta$ in all the three equations, hoping to square it and then obtain $\csc^2\theta$. Got stuck.

Then I tried extracting $\cos\theta$ from the third equation, squaring it and finding $\csc^2\theta$ and somehow manipulating the first two equations to arrive at $c^2+a^2$. That failed too.

The equations somewhat resemble the cosine rule, but I just can't figure out how to manipulate them with so many variables involved.

Please enlighten me.

Answer 1

(1) is written as $$(b + a\cos \theta)^2 = 1 - a^2 \sin^2\theta. \tag{4}$$

(2) is written as $$(d + c\cos \theta)^2 = 1 - c^2 \sin^2 \theta. \tag{5}$$

(3) is written as $$(b + a\cos \theta)(d + c\cos \theta) = - ac\sin^2\theta. \tag{6}$$

From (4)-(6), we have $$(1 - a^2\sin^2\theta)(1 - c^2 \sin^2\theta) = a^2c^2 \sin^4\theta$$ which yields $$c^2\sin^2\theta + a^2\sin^2\theta = 1.$$

We are done.