To prove $ \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) $ To prove  $ \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) $.
I tried to convert $\tan(4A)$ and $ \tan(8A)$ to $\tan(A$) and tried putting in L.H.S but it becomes a mess. Is there a easy and also intuitive  way to do this?
Thanks
 A: HINT:
$$\cot x-\tan x=2\cdot\frac{\cos^2x-\sin^2x}{2\sin x\cos x}=2\cot 2x$$
Set $x=A,2A,4A$   $$\cot A-\tan A=2\cot2A$$
$$2(\cot2A-\tan2A)=2(2\cot4A)$$
$$2^2(\cot4A-\tan4A)=2^2(2\cot8A)$$
Now add.
A: Derived formula -
cot A - tan A = $\frac 1{tan A} - \tan A = \frac{1 - \tan^2A}{tanA}$
= $\frac {2}{\frac{2\tan A}{1-\tan^2 A}} = \frac{2}{\tan 2A} = 2\cot 2A$
So, cot A - tan A = 2 cot 2A.
Now we have,
cot A – tan A – 2 tan 2A – 4 tan 4A – 8 cot 8A = 0
= 2 cot 2A – 2 tan 2A – 4tan 4A – 8 cot 8A
= 2(cot 2A – tan 2A) – 4 tan 4A – 8 cot 8 A
= 2 [2cot 2 (2A)] – 4 tan 4A – 8 cot 8A
= 4 cot 4A – 4 tan 4A – 8 cot 8A
= 4 (cot 4A – tan 4A) – 8 cot 8A
= 4 [2 cot 2 (4A)] – 8 cot 8A
= 8 cot 8A – 8 cot 8A
= 0
A: Add and subtract $\cot a$
$= \cot a + (\tan a - \cot a) + 2\tan 2a + 4 \tan 4a + 8 \cot 8a $
$= \cot a - 2\cot2a + 2\tan2a + 4 \tan 4a + 8 \cot 8a $
$= \cot a - 4 \cot4a + 4 \tan 4a + 8 \cot 8a $
$= \cot a - 8 \cot 8a + 8 \cot 8a $
$= \cot a$
A: $$\cos^6A-\sin^6A=\cos2A(1-1/4\sin^2(2A))$$
soln
$$=(\cos^2A)^3-(\sin^2A)^3$$
$$=(\cos^2A-\sin^2A)\big((\cos^2A+\sin^2A)^2-2\cos^2A\cdot \sin^2A+\cos^2A\cdot\sin^2A\big)$$
$$=\cos2A(1-\cos^2A\cdot\sin^2A)$$
$$=\cos2A(1-1/4(4\cos^2A\cdot\sin^2A)$$
$$=\cos2A\big(1-1/4\sin^2(2A)\big)$$
                                      RHS proved
                                                           #
