Question

If $ \tan \theta + \cot \theta = 4 $, then find the value of $ \tan^3 \theta + \cot^3 \theta $.

• A. 52
• B. 44
• C. 46
• D. 54

Hint:

To evaluate \( \tan^3 \theta + \cot^3 \theta \), use the identity: \[ a^3 + b^3 = (a + b)^3 - 3ab(a + b) \] Also, convert to a single variable using substitution like \( \tan \theta = x \Rightarrow \cot \theta = \frac{1}{x} \).

Answer 1

The Correct Option is C

Step 1: Use the identity for sum of cubes. We know: \[ \tan^3 \theta + \cot^3 \theta = (\tan \theta + \cot \theta)^3 - 3\tan \theta \cot \theta (\tan \theta + \cot \theta) \] Given: \[ \tan \theta + \cot \theta = 4 \] Let us compute: \[ (\tan \theta + \cot \theta)^3 = 4^3 = 64 \] Now, we need to find \( \tan \theta \cot \theta \). But: \[ \tan \theta \cot \theta = \tan \theta \cdot \frac{1}{\tan \theta} = 1 \] Step 2: Plug into the identity. \[ \tan^3 \theta + \cot^3 \theta = 64 - 3(1)(4) = 64 - 12 = 52 \] Step 3: Let \( x = \tan \theta \), then \( \cot \theta = \frac{1}{x} \). Given: \[ x + \frac{1}{x} = 4 \Rightarrow x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = 16 - 2 = 14 \] Now, \[ x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = 64 - 12 = 52 \] So: \[ \tan^3 \theta + \cot^3 \theta = 52 \] Correction: The correct answer is (A) 52