If `(1+tantheta)(1+tanphi)=2`, then `theta + phi`=

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `75^(@)`

Answer

To solve the problem where \((1 + \tan \theta)(1 + \tan \phi) = 2\), we will follow these steps:

### Step 1: Expand the left-hand side
We start by expanding the left-hand side of the equation:
\[ (1 + \tan \theta)(1 + \tan \phi) = 1 + \tan \theta + \tan \phi + \tan \theta \tan \phi \]
Thus, we can rewrite the equation as:
\[ 1 + \tan \theta + \tan \phi + \tan \theta \tan \phi = 2 \]

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

### Step 3: Use the tangent addition formula
We can use the tangent addition formula, which states:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
In our case, we can let \(A = \theta\) and \(B = \phi\). Thus, we can write:
\[ \frac{\tan \theta + \tan \phi}{1 - \tan \theta \tan \phi} = 1 \]

### Step 4: Set up the equation
From the previous step, we can set up the equation:
\[ \tan \theta + \tan \phi = 1 - \tan \theta \tan \phi \]

### Step 5: Substitute into the tangent addition formula
From the rearranged equation:
\[ \tan \theta + \tan \phi = 1 - \tan \theta \tan \phi \]
This means:
\[ \tan(\theta + \phi) = 1 \]

### Step 6: Solve for \(\theta + \phi\)
The equation \(\tan(\theta + \phi) = 1\) implies:
\[ \theta + \phi = 45^\circ + n \cdot 180^\circ \quad \text{for any integer } n \]
However, the principal value we are interested in is:
\[ \theta + \phi = 45^\circ \]

### Final Answer
Thus, the value of \(\theta + \phi\) is:
\[ \theta + \phi = 45^\circ \]

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