value of $\tan(A)$ What is value of $tanA$ if $2\tan(2A)+4\tan(4A)+\frac{8}{\tan(8A)}=0$ writing everything in $\tan(A)$ and solving for $t$ is next to impossible without maths engines. So i am seeking for a shorter way. Thanks
 A: First this equation is defined if and only if $2A\not\equiv \dfrac\pi2\mod\pi$, $\;4A\not\equiv \dfrac\pi2\mod\pi$ and $8A\not\equiv 0\mod\dfrac\pi2$, i.e. if and only if
$\begin{cases}A\not\equiv \dfrac\pi4,\dfrac \pi8\mod\dfrac\pi2,\\A\not\equiv 0\mod\dfrac\pi{16}.
\end{cases}$
The first condition is redundant w.r.t. the second, so it all  comes down to the simple
$$A\not\equiv 0\mod\dfrac\pi{16}.$$
Now use the duplication formulae for the tangent:
$$4\tan 4A+\frac8{\tan 8A}=4\tan 4A+ \frac{8}{\cfrac{2\tan4A}{1-\tan^24A}}=4\Bigl(\tan4A+\frac{1-\tan^24A}{\tan 4A}\Bigr)=\frac4{\tan 4A}$$ 
Similarly
$$2\tan 2A+4\tan 4A+\frac8{\tan 8A}=2\tan 2A+\frac4{\tan 4A}=\frac2{\tan2A},$$
which can't be $0$.  
However you might consider it means $\tan 2A=\pm\infty$, which means $2A\equiv \dfrac\pi2\mod\pi$, i.e. $A\equiv\dfrac\pi4\mod\dfrac\pi2$. 
A: Let $x=4A$.  Then, the equation $2\tan(2A)+4\tan(4A)+8\cot(8A)=0$ is equivalent to 
$$\tan(x/2)+2\tan(x)+4\cot(2x)=0 \tag 1$$
Using the double angle formula for the tangent function, we write 
$$\cot(2x)=\frac{1-\tan^2(x)}{2\tan(x)} \tag 2$$
Using $(2)$ we find that 
$$2\tan(x)+4\cot(2x)=2\cot(x) \tag 3$$
This reduces $(1)$ to the equation
$$\tan(x/2)+2\cot(x)=0 \tag 4$$
But comparing $(4)$ to $(3)$, we see that 
$$\tan(x/2)+2\cot(x)=\cot(x/2)$$
Therefore, we have reduced the problem to finding the roots of 
$$\cot(x/2)=0$$
These roots are at 
$$x=(2n+1)\pi$$
for all integer values of $n$.
Finally, we find that since $x=4A$, 
$$\bbox[5px,border:2px solid #C0A000]{A=(2n+1)\pi/4}$$

However, note that $\tan(2A)$ is not defined.  And $\cot(8A)$ is likewise not defined.  We have found spurious solutions that one might interpret as a solution in the sense of a limit or in the sense of the extended reals.

A: $\tan(\phi)+{2 \over \tan(2\phi)} = {\sin(\phi) \over \cos(\phi)} + {\cos(2\phi) \over \sin(\phi) \cos(\phi)} = {\sin^2(\phi) + \cos^2(\phi) - \sin^2(\phi) \over \sin(\phi) \cos(\phi)} = {\cos(\phi) \over \sin(\phi)} = {1 \over \tan(\phi)}$
$\tan(a)+2\tan(2a)+{4 \over \tan(4a)} = tan(a) + {2 \over \tan(2a)} = {1 \over \tan(a)} = 0$
A: HINT:
$$\cot A-\tan A=\cdots=2\cot2A$$
$$\implies2\tan2A+4\tan4A+8\cot8A$$
$$=2\cot2A-2(\cot2A-\tan2A)+4\tan4A+8\cot8A$$
$$=2\cot2A-2(2\cot4A)+4\tan4A+8\cot8A$$
$$=2\cot2A-8(\cot8A-\cot8A)$$
$$=2\cot2A$$
$$\implies\cot2A=0\iff\cos2A=0$$
$\implies2A$ must be odd multiple of $\dfrac\pi2$
