If `(1+tantheta)(1+tanphi)=2`, then what is `(theta+phi)` equal to ?

Options

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

Explanation

To solve the equation \((1 + \tan \theta)(1 + \tan \phi) = 2\), we can follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:
\[ (1 + \tan \theta)(1 + \tan \phi) = 1 + \tan \theta + \tan \phi + \tan \theta \tan \phi \]
This gives us:
\[ 1 + \tan \theta + \tan \phi + \tan \theta \tan \phi = 2 \]

### Step 2: Rearrange the equation
Next, we can rearrange the equation to isolate the terms involving \(\tan\):
\[ \tan \theta + \tan \phi + \tan \theta \tan \phi = 2 - 1 \]
This simplifies to:
\[ \tan \theta + \tan \phi + \tan \theta \tan \phi = 1 \]

### Step 3: Use the tangent addition formula
We can recognize that the left-hand side resembles the tangent addition formula:
\[ \tan(\theta + \phi) = \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} \]
From our rearranged equation, we can express it as:
\[ \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} = 1 \]

### Step 4: Set up the equation
This leads us to:
\[ \tan \theta + \tan \phi = 1 - \tan \theta \tan \phi \]
Since we already have \(\tan \theta + \tan \phi + \tan \theta \tan \phi = 1\), we can set:
\[ \tan(\theta + \phi) = 1 \]

### Step 5: Solve for \(\theta + \phi\)
The tangent function equals 1 at specific angles:
\[ \tan(\theta + \phi) = 1 \implies \theta + \phi = \frac{\pi}{4} \text{ or } 45^\circ \]

Thus, we conclude that:
\[ \theta + \phi = 45^\circ \]

### Final Answer
Therefore, the value of \((\theta + \phi)\) is \(45^\circ\).

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