To solve the given expression, we can use trigonometric identities and simplify the equation step by step. Here's a detailed explanation of the solution:

1. Start with the given expression: tan(alpha) * 2cot(2 alpha) * 4tan(4alpha) * 8cot(8alpha) = cot(alpha).

2. Let's simplify each term one by one:
a. Begin with the first term, tan(alpha). Using the identity tan(alpha) = sin(alpha)/cos(alpha), we can rewrite it as sin(alpha)/cos(alpha).
b. Moving on to the second term, 2cot(2 alpha). The cotangent identity cot(2 alpha) = cos(2 alpha)/sin(2 alpha) allows us to rewrite it as 2cos(2 alpha)/sin(2 alpha).
c. Now, let's simplify the third term, 4tan(4alpha). Using the tangent identity tan(4 alpha) = sin(4 alpha)/cos(4 alpha), we can rewrite it as 4sin(4 alpha)/cos(4 alpha).
d. Finally, we address the fourth term, 8cot(8alpha). Similar to the previous steps, we can rewrite it using the cotangent identity as 8cos(8 alpha)/sin(8 alpha).

3. Now, substitute these simplified expressions back into the original equation:
(sin(alpha)/cos(alpha)) * (2cos(2 alpha)/sin(2 alpha)) * (4sin(4 alpha)/cos(4 alpha)) * (8cos(8 alpha)/sin(8 alpha)) = cot(alpha).

4. Next, let's simplify the expression further:
a. Cancel out common factors between the numerator and denominator of each fraction. This gives us:
(sin(alpha) * 2cos(2 alpha) * 4sin(4 alpha) * 8cos(8 alpha)) / (cos(alpha) * sin(2 alpha) * cos(4 alpha) * sin(8 alpha)) = cot(alpha).

5. Now, let's simplify each term in the numerator and denominator:
a. In the numerator, we can use the double angle identities to simplify the trigonometric functions:
sin(2 alpha) = 2sin(alpha)cos(alpha)
cos(4 alpha) = 1 - 2sin^2(2 alpha)
sin(8 alpha) = 2sin(4 alpha)cos(4 alpha)

Substituting these identities into the numerator gives us:
(sin(alpha) * 2cos(2 alpha) * 4sin(4 alpha) * 8cos(8 alpha)) = (sin(alpha) * 2cos(2 alpha) * 4sin(4 alpha) * 8(1 - 2sin^2(2 alpha)) * 2sin(4 alpha) * (1 - 2sin^2(2 alpha)))

b. In the denominator, we can simplify the trigonometric functions:
sin(2 alpha) = 2sin(alpha)cos(alpha)

Substituting this identity into the denominator gives us:
(cos(alpha) * sin(2 alpha) * cos(4 alpha) * sin(8 alpha)) = (cos(alpha) * 2sin(alpha)cos(alpha) * (1 - 2sin

Answer tan2A=2tanA1−tan2Atan⁡2A=2tan⁡A1−tan2⁡A tanα+2tan2α+4tan4α+8cot8α=cotαtan⁡α+2tan⁡2α+4tan⁡4α+8cot⁡8α=cot⁡α (Or) cotα−tanα−2tan2α−4tan4α−8cot8α=0cot⁡α−tan⁡α−2tan⁡2α−4tan⁡4α−8cot⁡8α=0 cotα−tanα=1tanα−cot⁡α−tan⁡α=1tan⁡α−tanαtan⁡α ⇒1−tan2αtanα⇒1−tan2⁡αtan⁡α ⇒2tan2α⇒2tan⁡2α ⇒2cot2α⇒2cot⁡2α LHS 2cot2α−2tan2α−4tan4α−8cot8α2cot⁡2α−2tan⁡2α−4tan⁡4α−8cot⁡8α ⇒2(1tan2α−⇒2(1tan⁡2α−tan2α)−4tan4α−8cot8αtan⁡2α)−4tan⁡4α−8cot⁡8α ⇒21−tan22αtan2α⇒21−tan2⁡2αtan⁡2α−4tan4α−8cot8α−4tan⁡4α−8cot⁡8α ⇒4(1−tan22α)2tan2α⇒4(1−tan2⁡2α)2tan⁡2α−4tan4α−8cot8α−4tan⁡4α−8cot⁡8α ⇒4tan4α⇒4tan⁡4α−4tan4α−8cot8α−4tan⁡4α−8cot⁡8α ⇒4(1−tan24α)tan4α⇒4(1−tan2⁡4α)tan⁡4α−8cot8α−8cot⁡8α ⇒8(1−tan24α)2tan4α⇒8(1−tan2⁡4α)2tan⁡4α−8cot8α−8cot⁡8α ⇒8tan8α⇒8tan⁡8α−8cot8α−8cot⁡8α ⇒8cot8α−8cot8α⇒8cot⁡8α−8cot⁡8α=0=RHS Hence proved