If `0^@ le x le 90^@` and `1+tan^2 x - 2 tan x =0`, then the value of x is

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `90^@`

Answer

To solve the equation \(1 + \tan^2 x - 2 \tan x = 0\) for \(0^\circ \leq x \leq 90^\circ\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 1 + \tan^2 x - 2 \tan x = 0 \]
This can be rearranged to:
\[ \tan^2 x - 2 \tan x + 1 = 0 \]

### Step 2: Recognize the quadratic form
Notice that the equation is a quadratic in terms of \(\tan x\). We can let \(y = \tan x\), transforming our equation into:
\[ y^2 - 2y + 1 = 0 \]

### Step 3: Factor the quadratic
The quadratic can be factored as:
\[ (y - 1)^2 = 0 \]
This implies:
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]

### Step 4: Substitute back for \(\tan x\)
Since we let \(y = \tan x\), we have:
\[ \tan x = 1 \]

### Step 5: Solve for \(x\)
Now we need to find \(x\) such that:
\[ \tan x = 1 \]
In the interval \(0^\circ \leq x \leq 90^\circ\), the angle \(x\) that satisfies this is:
\[ x = 45^\circ \]

### Conclusion
Thus, the value of \(x\) is:
\[ \boxed{45^\circ} \]

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