Question

If Tan\(\theta\) + Cot\(\theta\) = 4, find Tan\(^2\theta\) + Cot\(^2\theta\).

• A. 14
• B. 16
• C. 20
• D. 8

Hint:

Whenever you see a question of the form "If \(x + 1/x = k\), find \(x^2 + 1/x^2\)", the answer is always \(k^2 - 2\). This problem is a direct application, since \(\cot(\theta) = 1/\tan(\theta)\). Here \(k=4\), so the answer is \(4^2 - 2 = 14\).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the sum of Tan\(\theta\) and Cot\(\theta\) and are asked to find the sum of their squares.
Step 2: Key Formula or Approach:
We will use the algebraic identity \((a + b)^2 = a^2 + 2ab + b^2\).
We also need the trigonometric identity \( \tan(\theta) \cdot \cot(\theta) = 1 \).
Step 3: Detailed Explanation:
Let's start with the given equation: \[ \tan(\theta) + \cot(\theta) = 4 \] To get the terms \(\tan^2\theta\) and \(\cot^2\theta\), we can square both sides of the equation.
\[ (\tan(\theta) + \cot(\theta))^2 = 4^2 \] Now, we expand the left side using the algebraic identity \((a+b)^2 = a^2 + 2ab + b^2\), where \(a = \tan(\theta)\) and \(b = \cot(\theta)\).
\[ \tan^2(\theta) + 2 \cdot \tan(\theta) \cdot \cot(\theta) + \cot^2(\theta) = 16 \] We know that \(\tan(\theta) \cdot \cot(\theta) = 1\), because \(\cot(\theta) = \frac{1}{\tan(\theta)}\).
Substitute this value into the equation: \[ \tan^2(\theta) + 2(1) + \cot^2(\theta) = 16 \] \[ \tan^2(\theta) + \cot^2(\theta) + 2 = 16 \] To find \(\tan^2(\theta) + \cot^2(\theta)\), we subtract 2 from both sides.
\[ \tan^2(\theta) + \cot^2(\theta) = 16 - 2 \] \[ \tan^2(\theta) + \cot^2(\theta) = 14 \] Step 4: Final Answer:
The value of \(\tan^2\theta + \cot^2\theta\) is 14.