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$$\tan A + \tan 2A + \tan 3A=0$$ I tried converting these all in sin and cos and I got the answer but the answer didn't match. So any one could just help me. Thanks

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Use this:

$\tan(3A)=\dfrac{\tan A+\tan 2A}{1-\tan A \tan 2A}$

Substitute the numerator with $-\tan 3A$ and you're going to get a expression like this:

$$\tan 3A(2-\tan A\tan 2A)=0$$

Meaning either $\tan 3A=0$ or $\tan A\tan 2A=2$.

$\tan 3A=0$ implies $A=\dfrac{n}{3}\pi$ where $n$ is an integer.

$\tan A\tan 2A=2$ means $\dfrac{2\tan^2 A}{1-\tan^2 A}=2$.

After simplifying, you're going to get

$$\tan A=\pm\frac{1}{\sqrt{2}}$$

You're going to get the other solutions from here.

I hope this helps.

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$$\tan A+\tan2A=\frac{\sin3A}{\cos A\cos2A}$$

$$\tan A+\tan2A+\tan3A=0\implies\frac{\sin3A}{\cos A\cos2A}+\frac{\sin3A}{\cos3A}=0$$

$$\iff\sin3A(\cos3A+\cos A\cos2A)=0$$

If $\sin3A=0,3A=n\pi$ where $n$ is any integer

Otherwise, $$\cos3A+\cos A\cos2A=0\iff 4\cos^3A-3\cos A+\cos A(2\cos^2A-1)=0$$

$$\iff \cos A(6\cos^2A-4)=0$$

If $\displaystyle\cos A=0, A=(2m+1)\frac\pi2$

Otherwise, $\displaystyle6\cos^2A-4=0\iff\cos^2A=\frac23,\cos2A=2\cos^2A-1=\frac13$

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Hint

The following identities for the tangent of multiple angles should be useful $$\tan(2A)=\frac{2\tan(A)}{1-2\tan^2(A)}$$ $$\tan(3A)=\frac{3\tan(A)-\tan^3(A)}{1-3\tan^2(A)}$$

I am sure that you can take from here.

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  • $\begingroup$ Actually I used these properties but the equation became tedious could you just please elaborate and explain it $\endgroup$ – Apoorv Jain Jul 20 '14 at 9:25
  • $\begingroup$ It is very simple. Change $\tan(A)$ to $x$; as you can see, $x$ can be factored. Reduce to same denominator and expand. You should end with a quadratic equation in $x^2$. $\endgroup$ – Claude Leibovici Jul 20 '14 at 9:29
  • $\begingroup$ Thats all nice just please update the answer $\endgroup$ – Apoorv Jain Jul 20 '14 at 9:31
  • $\begingroup$ I shall not update my answer repeating a previous comment. You did the work ! In turn, I suggest that you edit the solution in the post. It could be helpful to other participants to this forum. Cheers :) $\endgroup$ – Claude Leibovici Jul 20 '14 at 9:53

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