<s>[INST] 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.
 [/INST] 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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