On proving an identity given a system of trig equations We are given the following:
$$a^2 + b^2 + 2ab\cos\theta = 1 \tag1$$
$$d^2 + c^2 + 2cd\cos\theta = 1 \tag2$$
$$ac + bd + (ad + bc)\cos\theta = 0\tag3$$
It is required to prove that:
$$a^2 + c^2 = \csc^2\theta$$
$$b^2 + d^2 = \csc^2\theta$$
I right away noticed that this problem didn't change if we replaced $a, b$ with $c, d$ respectively. This suggested a that a reduction of the problem might be possible, but I couldn't quite put my finger on it. 
I then noticed that the first two equations represented the magnitude of the vector sum of the vectors $a, b$ and $c, d$. But again, this was not very helpful, as I couldn't relate the third equation to the problem.
I lastly tried to specialize the problem to a special case, hoping it would yield some useful insight. For instance, if $\cos\theta = 0$, the equations reduces to:
$$a^2 + b^2 = 1 \tag4$$
$$d^2 + c^2 = 1\tag5$$
$$ac + bd = 0\tag6$$
and we are required to prove::
$$a^2 + c^2 = b^2 + d^2 = 1$$
this can be immediately seen if we solve $(6)$ for $a$ and then substitute in $(4)$ getting, $b = c$. 
This however, did not give me any insight into solving the problem. In general, I have a lot of trouble figuring out how to relate the hypothesis given in such (algebraic) problems to the conclusion. Most of the time, I just keep transforming the equations aimlessly until I stumble upon the conclusion. 
I would prefer a solution with the motivation behind it, and also, if possible, some general tips on approaching such problems.
 A: Consider the map:
$f:(a,b)\to (a + b\cos\theta, b\sin\theta)$, the space being $\Bbb R^2$ with euclidian structure.
The hypothesis writes
$$
|f(a,b)|^2 = |f(c,d)|^2 = 1
\\
\langle f(a,b), f(c,d) \rangle = 0
$$
Now let us compute
$$
f^{-1}(x,y) = (x - {\cot\theta}y, y \csc\theta)$$
Consider a orthogonal basis $((x_1,y_1), (x_2,y_2))$ (under the hypothesis, $ (f(a,b), f(c,d))$ is one). You can find $\phi$ such as:
$$
(x_1,y_1) = (\cos\phi ,\sin\phi)
\\
(x_2,y_2) = (-\sin\phi ,\cos\phi)
\\
[f^{-1}(x_1,y_1)\cdot e_x]^2 + [f^{-1}(x_2,y_2)\cdot e_y]^2 = 
(\csc\phi - \cot\theta\sin\phi)^2 + 
(-\sin\phi- \cot\theta\cos\phi)^2 = \csc^2\phi 
$$
A: So you are given the following: 
$$a^2 + b^2 + 2ab\cos\theta = 1 \tag1$$
$$d^2 + c^2 + 2cd\cos\theta = 1 \tag2$$
$$ac + bd + (ad + bc)\cos\theta = 0\tag3$$
And you want to prove:
$$a^2 + c^2 = \csc^2\theta$$
$$b^2 + d^2 = \csc^2\theta$$
To Begin
As you stated above, you started by says that $$\cos(\theta) = 0$$
Then 
$$a^2 + b^2 = 1 \tag1$$
$$d^2 + c^2 = 1 \tag2$$
$$ac + bd = 0\tag3$$
Realize that if $\cos(\theta) = 0$, then $\csc^2(\theta) = 1$ 
Therefore 
$$a^2 + c^2 = 1$$
$$b^2 + d^2 = 1$$
If you add these 2 equation together 
$$a^2 + b^2 + c^2 + d^2 = 2$$
Adding (1) and (2) gives us
$$a^2 + b^2 + c^2 + d^2 = 2$$
And thus it is proven
