`tan 70^(@) - tan 20^(@) =`

Options

A. `tan 50^(@)`
B. `cot 50^(@)`
C. `2 tan 50^(@)`
D. `2 cot 50^(@)`

Answer

To solve the expression \( \tan 70^\circ - \tan 20^\circ \), we can use the tangent subtraction formula. The formula states that:

\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

However, in our case, we can also use the identity for the tangent of complementary angles. Notably, we have:

\[ \tan(90^\circ - x) = \cot x \]

This means that:

\[ \tan 70^\circ = \cot 20^\circ \]

Now, we can rewrite the expression:

1. **Substituting the identity:**
\[ \tan 70^\circ - \tan 20^\circ = \cot 20^\circ - \tan 20^\circ \]

2. **Using the identity for cotangent:**
\[ \cot 20^\circ = \frac{1}{\tan 20^\circ} \]

3. **Substituting this back into the expression:**
\[ \cot 20^\circ - \tan 20^\circ = \frac{1}{\tan 20^\circ} - \tan 20^\circ \]

4. **Finding a common denominator:**
\[ = \frac{1 - \tan^2 20^\circ}{\tan 20^\circ} \]

5. **Using the Pythagorean identity:**
Since \(1 - \tan^2 x = \frac{1}{\sec^2 x}\), we can rewrite it as:
\[ 1 - \tan^2 20^\circ = \frac{1}{\sec^2 20^\circ} \]

6. **Substituting this back:**
\[ = \frac{\frac{1}{\sec^2 20^\circ}}{\tan 20^\circ} = \frac{1}{\tan 20^\circ \sec^2 20^\circ} \]

7. **Using the identity \(\sec^2 x = 1 + \tan^2 x\):**
\[ = \frac{1}{\tan 20^\circ (1 + \tan^2 20^\circ)} \]

8. **Final simplification:**
This expression can be simplified further, but we can also evaluate it numerically or use a calculator to find the approximate value.

Thus, the final answer is:

\[ \tan 70^\circ - \tan 20^\circ = 2 \]