If `A +B = 45 ^(@),` show that `tan A + tan B + tan A tan B =1.` Hence or otherwise, express `tan 22 ^(@) 30'` in surd form.

Answer

To solve the problem step by step, we will first prove that if \( A + B = 45^\circ \), then \( \tan A + \tan B + \tan A \tan B = 1 \). Then, we will express \( \tan 22^\circ 30' \) in surd form.

### Step 1: Start with the given equation
We are given that:
\[ A + B = 45^\circ \]

### Step 2: Apply the tangent function
Taking the tangent of both sides, we have:
\[ \tan(A + B) = \tan(45^\circ) \]
Since \( \tan(45^\circ) = 1 \), we can write:
\[ \tan(A + B) = 1 \]

### Step 3: Use the tangent addition formula
Using the tangent addition formula:
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]
We can set this equal to 1:
\[ \frac{\tan A + \tan B}{1 - \tan A \tan B} = 1 \]

### Step 4: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:
\[ \tan A + \tan B = 1 - \tan A \tan B \]

### Step 5: Rearranging the equation
Rearranging the equation:
\[ \tan A + \tan B + \tan A \tan B = 1 \]

### Conclusion of the proof
Thus, we have shown that:
\[ \tan A + \tan B + \tan A \tan B = 1 \]

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### Step 6: Express \( \tan 22^\circ 30' \) in surd form
Now we need to express \( \tan 22^\circ 30' \) in surd form. We can convert \( 22^\circ 30' \) into decimal degrees:
\[ 22^\circ 30' = 22.5^\circ \]
Let \( \theta = 22.5^\circ \). Then \( 2\theta = 45^\circ \).

### Step 7: Use the double angle formula for tangent
Using the double angle formula:
\[ \tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta} \]
Since \( \tan(45^\circ) = 1 \), we have:
\[ \frac{2\tan \theta}{1 - \tan^2 \theta} = 1 \]

### Step 8: Cross-multiply and rearrange
Cross-multiplying gives:
\[ 2\tan \theta = 1 - \tan^2 \theta \]
Rearranging this results in:
\[ \tan^2 \theta + 2\tan \theta - 1 = 0 \]

### Step 9: Solve the quadratic equation
Using the quadratic formula \( \tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = 2, c = -1 \):
\[ \tan \theta = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ = \frac{-2 \pm \sqrt{4 + 4}}{2} \]
\[ = \frac{-2 \pm \sqrt{8}}{2} \]
\[ = \frac{-2 \pm 2\sqrt{2}}{2} \]
\[ = -1 \pm \sqrt{2} \]

### Step 10: Choose the appropriate solution
Since \( \theta = 22.5^\circ \) is in the first quadrant, we take the positive root:
\[ \tan 22.5^\circ = \sqrt{2} - 1 \]

### Final Answer
Thus, we express \( \tan 22^\circ 30' \) in surd form as:
\[ \tan 22^\circ 30' = \sqrt{2} - 1 \]

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