Question

If \(\tan A - \frac{1}{\tan A} = 1\). Then value of \(\tan^2 A + \frac{1}{\tan^2 A}\) is :

• A. 3
• B. 1
• C. 2
• D. -2

Hint:

Let \(y = \tan A\). The problem gives \(y - \frac{1}{y} = 1\). We need to find \(y^2 + \frac{1}{y^2}\). 1. Square the given equation: \(\left(y - \frac{1}{y}\right)^2 = 1^2\). 2. Expand the left side: \(y^2 - 2 \cdot y \cdot \frac{1}{y} + \left(\frac{1}{y}\right)^2 = 1\). 3. Simplify: \(y^2 - 2 + \frac{1}{y^2} = 1\). 4. Isolate the desired term: \(y^2 + \frac{1}{y^2} = 1 + 2 = 3\). So, \(\tan^2 A + \frac{1}{\tan^2 A} = 3\).

Answer 1

The Correct Option is A

Concept: This problem uses an algebraic manipulation similar to the identity \((x-y)^2 = x^2 - 2xy + y^2\). We are given an expression involving a term and its reciprocal, and we need to find the sum of their squares. Step 1: Let \(x = \tan A\) The given equation can be written as: \[ x - \frac{1}{x} = 1 \] We need to find the value of \(x^2 + \frac{1}{x^2}\). Step 2: Square both sides of the given equation Square the equation \(x - \frac{1}{x} = 1\): \[ \left(x - \frac{1}{x}\right)^2 = (1)^2 \] Expand the left side using the identity \((a-b)^2 = a^2 - 2ab + b^2\), where \(a=x\) and \(b=\frac{1}{x}\): \[ x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 1 \] Step 3: Simplify the expanded equation \[ x^2 - 2(1) + \frac{1}{x^2} = 1 \] \[ x^2 - 2 + \frac{1}{x^2} = 1 \] Step 4: Solve for \(x^2 + \frac{1}{x^2}\) To isolate \(x^2 + \frac{1}{x^2}\), add 2 to both sides of the equation: \[ x^2 + \frac{1}{x^2} = 1 + 2 \] \[ x^2 + \frac{1}{x^2} = 3 \] Step 5: Substitute back \(\tan A\) for \(x\) Since we let \(x = \tan A\), then \(x^2 + \frac{1}{x^2} = \tan^2 A + \frac{1}{\tan^2 A}\). Therefore, \[ \tan^2 A + \frac{1}{\tan^2 A} = 3 \] The value of \(\tan^2 A + \frac{1}{\tan^2 A}\) is 3. This matches option (1).