Question

If \(\text{tan} \theta + \frac{1}{\text{tan} \theta}= 2\), then the value of \(\text{tan}^2 \theta + \frac{1}{\text{tan}^2 \theta}\) is:

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

This type of problem often uses the algebraic identity \[ (a+b)^2 = a^2 + b^2 + 2ab. \] Alternatively, for expressions like \[ x + \frac{1}{x} = k, \] recognize that squaring both sides gives \[ x^2 + \frac{1}{x^2} = k^2 - 2. \] For the specific case where \[ x + \frac{1}{x} = 2, \] it implies \[ x=1 \] (since \[ x^2 - 2x + 1 = 0 \implies (x-1)^2 = 0). \]

Answer 1

The Correct Option is A

Step 1: Identify the given equation and the expression to be found. 
Given equation: $\text{tan} \theta + \frac{1}{\text{tan} \theta} = 2$ Expression to find: $\text{tan}^2 \theta + \frac{1}{\text{tan}^2 \theta}$ Step 2: Use an algebraic identity to relate the given and required expressions. 
This problem can be solved by squaring the given equation.
Let $x = \text{tan} \theta$. Then the given equation is $x + \frac{1}{x} = 2$.
We need to find $x^2 + \frac{1}{x^2}$.
Recall the algebraic identity: $(a+b)^2 = a^2 + b^2 + 2ab$.
In our case, let $a = x$ and $b = \frac{1}{x}$.
So, $\left(x + \frac{1}{x}\right)^2 = x^2 + \left(\frac{1}{x}\right)^2 + 2 \left(x\right) \left(\frac{1}{x}\right)$
$\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2$
Step 3: Substitute the given value into the identity. 
We know that $x + \frac{1}{x} = 2$.
Substitute this into the squared identity:
$(2)^2 = x^2 + \frac{1}{x^2} + 2$
$4 = x^2 + \frac{1}{x^2} + 2$
Step 4: Solve for the required expression. 
Rearrange the equation to find $x^2 + \frac{1}{x^2}$:
$x^2 + \frac{1}{x^2} = 4 - 2$
$x^2 + \frac{1}{x^2} = 2$
Since $x = \text{tan} \theta$, we have $\text{tan}^2 \theta + \frac{1}{\text{tan}^2 \theta} = 2$.
Alternative Method (Directly Solving for tan $\theta$):
Step 1: Given $\text{tan} \theta + \frac{1}{\text{tan} \theta} = 2$.
Let $y = \text{tan} \theta$.
$y + \frac{1}{y} = 2$
Multiply by $y$ to clear the denominator:
$y^2 + 1 = 2y$
$y^2 - 2y + 1 = 0$
Step 2: Factor the quadratic equation.
This is a perfect square trinomial: $(y-1)^2 = 0$.
So, $y - 1 = 0 \implies y = 1$.
Therefore, $\text{tan} \theta = 1$.
Step 3: Substitute the value of $\text{tan} \theta$ into the required expression.
$\text{tan}^2 \theta + \frac{1}{\text{tan}^2 \theta} = (1)^2 + \frac{1}{(1)^2}$
$= 1 + \frac{1}{1}$
$= 1 + 1 = 2$.
Both methods yield the same result.
Step 5: Final Answer.
The value of $\text{tan}^2 \theta + \frac{1}{\text{tan}^2 \theta}$ is 2.