# Need help to solve the trigonometric equation $\tan\frac {1+x}{2} \tan \frac {1+x}{3}=-1$

How can I solve the equation $\tan\frac {1+x}{2} \tan \frac {1+x}{3}=-1$. Please give me some hint for that. Thank you.

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Did you mean $\tan(\frac{1+x}2)?$ –  lab bhattacharjee Jan 9 '13 at 8:45
Oh sorry. I fixed it. –  Ned Dabby Jan 9 '13 at 8:47

Assuming $\frac12,\frac13$ are measured in radian,

$$\tan\left(\frac{1+ x}2\right)=-\frac1{\tan\left(\frac{1+ x}3\right)}=-\cot\left(\frac{1+ x}3\right)=\tan\left(\frac\pi2+\frac{x+1}3\right)$$ as $\tan(\frac\pi2+C)=-\cot C$

So, $$\frac{1+ x}2=n\pi+\frac\pi2+\frac{x+1}3$$ where $n$ is any integer as $\tan A=\tan B\implies A=n\pi+B$.

Alternatively, we know $$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$

If $\tan A\tan B=-1,\tan(A-B)=\infty\implies A-B=m\pi+\frac\pi2$ where $m$ is any integer

or, $$\tan A\tan B=-1\implies \frac{\sin A}{\cos A}=-\frac{\cos B}{\sin B}$$ $$\implies \cos A\cos B+\sin A\sin B=0\implies \cos(A-B)=0\implies A-B=(2r+1)\frac\pi2$$ where $r$ is any integer.

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Indeed, I was stuck in what you stated thorugh the last paragraph. –  Ned Dabby Jan 9 '13 at 8:58
@NedDabby, hope, I could make the idea clear. Added another method to avoid dealing with $\infty$ –  lab bhattacharjee Jan 9 '13 at 9:16
Hint: According to your equation, note that $\tan\left(1+\frac{x}{2}\right)=-\cot\left(1+\frac{x}{3}\right)=\tan\left(\frac{\pi}{2}+1+\frac{x}{3}\right)$
Should be $\frac {x}{3}$ at the end? –  Calvin Lin Jan 9 '13 at 8:57