Question

If $\tan A+2\tan 2A+4\tan 4A+8\cot 8A=k\cot A$, then find the value of $k$.

Answer 1

As $\cot A-\tan A=\frac{\cos A}{\sin A}-\frac{\sin A}{\cos A}=\frac{{\cos }^{2}A-{\sin }^{2}A}{\sin A\cos A}=2\cot 2A$

$\Rightarrow \tan A=\cot A-2\cot 2A$

Our expression $L.H.S$ becomes

$\cot A-2\cot 2A+2\tan 2A+4\tan 4A+8\cot 8A$

$=\cot A-2(\cot 2A-\tan 2A)+4\tan 4A+8\cot 8A$

$=\cot A-2\left(2\cot 4A\right)+4\tan 4A+8\cot 8A$

$=\cot A-4(\cot 4A-\tan 4A)+8\cot 8A$

$=\cot A-8\cot 8A+8\cot 8A$

$=\cot A\Rightarrow k=1$