Question

If $ \tan\left( \frac{\pi}{4} + \alpha \right) = \tan^3\left( \frac{\pi}{4} + \beta \right) $, then compute: $$ \tan(\alpha + \beta) \cot(\alpha - \beta) = ? $$

• A. \( \sec^2 2\beta + \tan^2 2\beta \)
• B. \( \csc^2 2\beta + \cot^2 2\beta \)
• C. \( 2(\sec^2 2\beta + \tan^2 2\beta) \)
• D. \( 4(\sec^2 2\beta + \tan^2 2\beta) \)

Hint:

Use identities: \( \tan(P+Q)\cot(P-Q) = \cot(P+Q)\cot(P-Q) \) and express in terms of \( \tan A, \tan B \) to simplify.

Answer 1

The Correct Option is C

Given: \[ \tan\left( \frac{\pi}{4} + \alpha \right) = \tan^3\left( \frac{\pi}{4} + \beta \right) \Rightarrow \tan A = \tan^3 B \text{ where } A = \frac{\pi}{4} + \alpha,\ B = \frac{\pi}{4} + \beta \] Let: \[ \tan A = t,\ \Rightarrow \tan B = \sqrt[3]{t} \] Then: \[ \alpha = A - \frac{\pi}{4},\quad \beta = B - \frac{\pi}{4} \Rightarrow \alpha + \beta = A + B - \frac{\pi}{2},\quad \alpha - \beta = A - B \] Now: \[ \tan(\alpha + \beta) \cot(\alpha - \beta) = \tan(A + B - \frac{\pi}{2}) \cot(A - B) = \cot(A + B) \cot(A - B) \Rightarrow \cot(A + B) \cot(A - B) = \frac{1 + \tan A \tan B}{\tan A - \tan B} \cdot \frac{1 - \tan A \tan B}{\tan A + \tan B} \] Use identity: \[ \cot(P + Q)\cot(P - Q) = \frac{1 + \tan P \tan Q}{\tan P - \tan Q} \cdot \frac{1 - \tan P \tan Q}{\tan P + \tan Q} = \frac{1 - (\tan A \tan B)^2}{\tan^2 A - \tan^2 B} \] Since \( \tan A = t,\ \tan B = t^{1/3} \), plug and simplify: After simplification, we obtain: \[ \tan(\alpha + \beta)\cot(\alpha - \beta) = 2(\sec^2 2\beta + \tan^2 2\beta) \]