Help me solve a trigonometric equation 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.
 A: 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.
A: Let's rewrite your equation :
$$\frac {\sin(t_1)+\cos(t_1)}{\sin(t_1)\cos(t_1)} =\frac {\sin(t_2)+\cos(t_2)}{\sin(t_2)\cos(t_2)}$$
that you may multiply by $\frac 12\sin\left(\frac{\pi}4\right)$ and rewrite as :
$$\frac {\sin\left(t_1+\frac {\pi}4\right)}{\sin(2\;t_1)} =\frac {\sin\left(t_2+\frac {\pi}4\right)}{\sin(2\;t_2)}$$
We could 'linearize' this to obtain :
$$\sin\left(t_1+\frac {\pi}4\right)\sin(2\;t_2)=\sin\left(t_2+\frac {\pi}4\right)\sin(2\;t_1)$$
But it will be more convenient to set $\;x_1:=t_1+\frac {\pi}4,\ x_2:=t_2+\frac {\pi}4\;$ and get (since $\sin\bigl(2x-\frac {\pi}2\bigr)=-\cos(2x)$ and reverting the fractions) :
$$\frac{\cos(2\;x_2)}{\sin(x_2)}=\frac{\cos(2\;x_1)}{\sin(x_1)}$$
and rewrite :
$$\frac{\cos(2\;x)}{\sin(x)}=\frac {1-2\sin(x)^2}{\sin(x)}=\frac 1{\sin(x)}-2\;\sin(x)$$
Since the solutions of $\frac 1s-2s=r$ are obtained by resolving $2s^2+rs-1=0$ with solutions $\;s=\frac{-r\pm\sqrt{8+r^2}}4$ we may deduce, for $s=\sin(x_2)$ and $r=\frac{\cos(2\;x_1)}{\sin(x_1)}$, that :
\begin{align}
\sin(x_2)&=\frac{-\cos(2\;x_1)\pm\sqrt{8\;\sin(x_1)^2+\cos(2\;x_1)^2}}{4\;\sin(x_1)}\\
&=\frac{-\cos(2\;x_1)\pm\left(2-\cos(2\;x_1)\right)}{4\;\sin(x_1)}\\
\end{align}
With the two different solutions :
\begin{align}
\sin(x_2)&=\frac{1-\cos(2\;x_1)}{2\;\sin(x_1)}=\sin(x_1)\\
\sin(x_2)&=-\frac{1}{2\;\sin(x_1)}\\
\end{align}
From this we deduce the four different solutions (modulo $2\pi$) :
\begin{align}
x_2&=x_1\\
x_2&=-\arcsin\left(\frac{1}{2\;\sin(x_1)}\right)\\
x_2&=\pi-x_1\\
x_2&=\pi+\arcsin\left(\frac{1}{2\;\sin(x_1)}\right)\\
\end{align}
The substitution $t_1=x_1-\frac {\pi}4$ and $t_2=x_2-\frac {\pi}4$ should give you the wished solutions.
