Find the value of `tantheta(1+sec2theta)(1+sec4theta)(1+sec8theta)`, when `theta=pi/32`

Answer

To find the value of \( \tan \theta (1 + \sec^2 \theta)(1 + \sec^4 \theta)(1 + \sec^8 \theta) \) when \( \theta = \frac{\pi}{32} \), we can follow these steps:

### Step 1: Write down the expression
We need to evaluate:
\[ \tan \theta (1 + \sec^2 \theta)(1 + \sec^4 \theta)(1 + \sec^8 \theta) \]

### Step 2: Substitute \( \theta = \frac{\pi}{32} \)
First, we need to find \( \tan \frac{\pi}{32} \) and the secant values:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Thus, we can express \( \sec^2 \theta \) in terms of \( \tan \theta \).

### Step 3: Calculate \( 1 + \sec^2 \theta \)
Using the identity:
\[ 1 + \sec^2 \theta = 1 + \frac{1}{\cos^2 \theta} = \frac{\cos^2 \theta + 1}{\cos^2 \theta} \]
This simplifies to:
\[ 1 + \sec^2 \theta = \frac{1 + \tan^2 \theta}{\cos^2 \theta} \]

### Step 4: Calculate \( 1 + \sec^4 \theta \)
Using the identity:
\[ 1 + \sec^4 \theta = (1 + \sec^2 \theta)^2 - 2 \]
This can be expressed in terms of \( \tan^2 \theta \):
\[ 1 + \sec^4 \theta = \left(1 + \tan^2 \theta\right)^2 - 2 = \tan^4 \theta + 2\tan^2 \theta \]

### Step 5: Calculate \( 1 + \sec^8 \theta \)
Similarly, we can write:
\[ 1 + \sec^8 \theta = (1 + \sec^4 \theta)^2 - 2 \]
This can also be expressed in terms of \( \tan^2 \theta \).

### Step 6: Combine the expressions
Now we can combine these expressions together:
\[ \tan \theta (1 + \sec^2 \theta)(1 + \sec^4 \theta)(1 + \sec^8 \theta) \]
Each of these terms can be simplified to yield a final expression involving \( \tan \theta \).

### Step 7: Evaluate \( \tan 8\theta \)
Using the double angle formulas:
\[ \tan 2\theta = \frac{2\tan \theta}{1 - \tan^2 \theta} \]
Continuing this process, we can find \( \tan 8\theta \) when \( \theta = \frac{\pi}{32} \):
\[ \tan 8\theta = \tan \left( \frac{8\pi}{32} \right) = \tan \left( \frac{\pi}{4} \right) = 1 \]

### Final Answer
Thus, the value of \( \tan \theta (1 + \sec^2 \theta)(1 + \sec^4 \theta)(1 + \sec^8 \theta) \) when \( \theta = \frac{\pi}{32} \) is:
\[ \boxed{1} \]