To prove $ \tan(A) + 2 \tan(2A) +4\tan4A + 8 \cot8A =\cot(A) $

Solution 1:

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.

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