Given:tan A =√7/21, 0°<A<90°  Find:tan 2A

Given that \(\tan A = \frac{\sqrt{7}}{21}\) and \(0^\circ < A < 90^\circ\), we need to find \(\tan 2A\).

### Step 1: Use the Double Angle Formula for Tangent

The double angle formula for tangent is:

\[
\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}
\]

### Step 2: Substitute the Value of \(\tan A\)

Substituting \(\tan A = \frac{\sqrt{7}}{21}\) into the formula:

\[
\tan 2A = \frac{2 \left(\frac{\sqrt{7}}{21}\right)}{1 - \left(\frac{\sqrt{7}}{21}\right)^2}
\]

### Step 3: Simplify the Expression

First, simplify the numerator:

\[
2 \times \frac{\sqrt{7}}{21} = \frac{2\sqrt{7}}{21}
\]

Now, simplify the denominator:

\[
\left(\frac{\sqrt{7}}{21}\right)^2 = \frac{7}{441}
\]

Thus, the denominator becomes:

\[
1 - \frac{7}{441} = \frac{441 - 7}{441} = \frac{434}{441}
\]

So the expression for \(\tan 2A\) now is:

\[
\tan 2A = \frac{\frac{2\sqrt{7}}{21}}{\frac{434}{441}}
\]

Simplify this by multiplying by the reciprocal:

\[
\tan 2A = \frac{2\sqrt{7}}{21} \times \frac{441}{434} = \frac{2\sqrt{7} \times 441}{21 \times 434}
\]

Simplify further:

\[
\tan 2A = \frac{2\sqrt{7} \times 21}{434} = \frac{42\sqrt{7}}{434}
\]

Simplify by dividing by the greatest common divisor of 42 and 434:

\[
\tan 2A = \frac{21\sqrt{7}}{217}
\]

So, the value of \(\tan 2A\) is:

\[
\tan 2A = \frac{21\sqrt{7}}{217}
\]

Would you like more details, or do you have any questions?

### Related Questions:

1. How would you find \(\tan 3A\) given \(\tan A\)?
2. Can you solve for \(\sin 2A\) and \(\cos 2A\) given \(\tan A\)?
3. What is the double angle identity for cosine and sine?
4. How do you convert tangent values into angle measures?
5. What is the general approach to simplifying trigonometric expressions?

**Tip:** Always remember to simplify the expression step-by-step to avoid errors, especially when dealing with fractions.

Given: tan A = √7/21, 0° < A < 90°, Find: tan 2A

Learn how to find tan 2A given tan A = √7/21 using the double angle formula for tangent. This detailed step-by-step solution walks through trigonometric identities and simplifies the expression, making it suitable for high school students in Grades 10-12.

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