Question

If \(\tan \theta + \cot \theta = 5\),then the value of \(\tan^2 \theta + \cot^2 \theta\) is

• A. 1
• B. 7
• C. 23
• D. 25

Answer 1

The Correct Option is C

Given: \( \tan \theta + \cot \theta = 5 \) 

Step 1: Express in terms of tan

\[ \cot \theta = \frac{1}{\tan \theta} \] So, \[ \tan \theta + \frac{1}{\tan \theta} = 5 \]

Step 2: Square both sides

\[ \left(\tan \theta + \frac{1}{\tan \theta} \right)^2 = 5^2 \] Expanding using the identity \( (a + b)^2 = a^2 + b^2 + 2ab \), \[ \tan^2 \theta + \cot^2 \theta + 2 = 25 \]

Step 3: Solve for \( \tan^2 \theta + \cot^2 \theta \)

\[ \tan^2 \theta + \cot^2 \theta = 25 - 2 = 23 \]

Final Answer: 23

Answer 2

Given the equation \(\tan \theta + \cot \theta = 5\), we are tasked with finding the value of \(\tan^2 \theta + \cot^2 \theta\).

Let's denote \(\tan \theta\) by \(x\).
Therefore, \(\cot \theta = \frac{1}{x}\).

The equation becomes \(x + \frac{1}{x} = 5\).

To find \(\tan^2 \theta + \cot^2 \theta\), calculate \((x^2 + \frac{1}{x^2})\) using the identity:
\((x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}\)

Substituting \(x + \frac{1}{x} = 5\) yields:
\((5)^2 = x^2 + 2 + \frac{1}{x^2}\)

Solving for \(x^2 + \frac{1}{x^2}\):
\(25 = x^2 + 2 + \frac{1}{x^2}\)
\(x^2 + \frac{1}{x^2} = 25 - 2 = 23\)

Therefore, the value of \(\tan^2 \theta + \cot^2 \theta\) is \(\boxed{23}\).