TITLE: Help me solve a trigonometric equation
QUESTION [5 upvotes]: I am doing some work in RF circuit design. I need to solve an equation for my design:
$$\frac 1{\cos(t_1)}+\frac 1{\sin(t_1)} =\frac 1{\cos(t_2)}+\frac 1{\sin(t_2)}$$
(I created a nicely typed image of this equation but setting don't allow me to upload unless I get at least 10 !]
Kindly help me find a general solution of the above equation (i.e. to find relation between $t_2$ and $t_1$). I may solve easier equations (like $\,\cos t=0.5$ !), but seems complex to me.
I observe that at $t_1=t_2=\dfrac {\pi}4$, this equation holds true. But I am unable to proceed further.
Thanks.
Mohammad A Maktoomi.

REPLY [7 votes]: Put $\sin\bigl(t+{\pi\over4}\bigr)=:u$. Then
$${1\over\cos t}+{1\over\sin t}={\sqrt{2}\sin\bigl(t+{\pi\over4}\bigr)\over{1\over2}\sin(2t)}=-{2\sqrt{2} u\over\cos\bigl(2(t+{\pi\over4})\bigr)}=-2\sqrt{2}{u\over 1-2u^2}\ .$$
Disregarding the cases $u=\pm{1\over\sqrt{2}}$ we therefore have to consider the equation
$${u_1\over 1-2u_1^2}={u_2\over 1-2u_2^2}\ ,$$
or 
$$(u_1-u_2)(1+2u_1u_2)=0\ .$$
There is the obvious solution $u_1=u_2$, which leads to (i) $t_1=t_2$ or  (ii) $t_1+t_2={\pi\over2}$.
The solutions to $1+2u_1u_2=0$ can be parametrized by
$$u_1=\pm {1\over\sqrt{2}} e^\tau\ ,\quad u_2=\mp {1\over\sqrt{2}} e^{-\tau}\qquad(-\log\sqrt{2}\leq \tau\leq\log\sqrt{2})\tag{1}$$
(note that $u_1$ and $u_2$ are sines). Each pair $(u_1,u_2)$ produced by $(1)$ will lead to two pairs $(t_1,t_2)$ solving the original equation; see the cases (i) and (ii) above.