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.

  • 1
    $\begingroup$ Hello, welcome to Math.SE. You can use LaTeX syntax on here. For some basic information about writing maths at this site see e.g. here, here, here and here. $\endgroup$
    – Lord_Farin
    May 14, 2013 at 11:23
  • $\begingroup$ A solution is $t_2=\pi/2-t_1$. Other solutions are extremely complicated. $\endgroup$ May 14, 2013 at 11:30
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    $\begingroup$ Raymond, Thanks for editing the question. Lord_Frin, Thanks for your links, i will go through them. Tavares, thank you- I will design with this solution. If this doesn't work, i will get back here. Thanks all, for your quick response. $\endgroup$ May 14, 2013 at 11:37
  • $\begingroup$ You are welcome ! Seeing my edit you will notice that this is not so much work... $\endgroup$ May 14, 2013 at 11:41

2 Answers 2


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.


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.

  • $\begingroup$ Thanks Christian and Raymond for your kind help. Actually, special solution due to Tavares didn't fit into my design.I will try to use these general solution. Hopefully these should work. Thanks, once again. $\endgroup$ May 16, 2013 at 12:03
  • $\begingroup$ You are welcome @MdAyatullahMaktoom. To clarify let's add that the other solutions, found by Christian first, came from : $$\sin\left(t_1+\frac{\pi}4\right)\sin\left(t_2+\frac{\pi}4\right)=-\frac 12$$ $\endgroup$ May 16, 2013 at 12:42

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