Solving the Trigonometry Problem: Evaluating tan(2 tan⁻¹(1/3))

The question asks us to find the value of the trigonometric expression \( \tan \left\{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right\} \). This involves evaluating an expression containing both the tangent function and the inverse tangent function.

Understanding Inverse Tangent and Related Formulas

The term \( {{\tan }^{-1}}\left( \frac{1}{3} \right) \) represents an angle whose tangent is \( \frac{1}{3} \). We need to evaluate the tangent of twice this angle.

A key formula relating \( 2{{\tan }^{-1}}(x) \) to the inverse tangent function is:

\( 2{{\tan }^{-1}}(x) = {{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right) \), provided \( |x| < 1 \).

In our problem, \( x = \frac{1}{3} \). Since \( \left| \frac{1}{3} \right| = \frac{1}{3} \), and \( \frac{1}{3} < 1 \), this formula is applicable.

Step-by-Step Calculation of tan(2 tan⁻¹(1/3))

Let's first evaluate the expression inside the tangent function, \( 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \), using the formula mentioned above.

Substitute \( x = \frac{1}{3} \) into the formula \( 2{{\tan }^{-1}}(x) = {{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right) \):

\( 2{{\tan }^{-1}}\left( \frac{1}{3} \right) = {{\tan }^{-1}}\left( \frac{2 \times \frac{1}{3}}{1-{{\left( \frac{1}{3} \right)}^{2}}} \right) \)

Now, simplify the expression inside the \( {{\tan }^{-1}} \) function:

\( \frac{2 \times \frac{1}{3}}{1-{{\left( \frac{1}{3} \right)}^{2}}} = \frac{\frac{2}{3}}{1-\frac{1}{9}} \)

Find a common denominator for the terms in the denominator:

\( 1-\frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{9-1}{9} = \frac{8}{9} \)

So, the expression becomes:

\( \frac{\frac{2}{3}}{\frac{8}{9}} \)

To divide fractions, multiply the numerator by the reciprocal of the denominator:

\( \frac{2}{3} \times \frac{9}{8} \)

Cancel common factors (3 into 9, and 2 into 8):

\( \frac{^{\cancel{2}1}}{_{\cancel{3}1}} \times \frac{^{\cancel{9}3}}{_{\cancel{8}4}} = \frac{1 \times 3}{1 \times 4} = \frac{3}{4} \)

Therefore, we have:

\( 2{{\tan }^{-1}}\left( \frac{1}{3} \right) = {{\tan }^{-1}}\left( \frac{3}{4} \right) \)

Now, substitute this back into the original expression we need to evaluate:

\( \tan \left\{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right\} = \tan \left\{ {{\tan }^{-1}}\left( \frac{3}{4} \right) \right\} \)

The property of inverse trigonometric functions states that \( \tan({{\tan }^{-1}}(y)) = y \) for any real number \( y \). Here, \( y = \frac{3}{4} \).

So, \( \tan \left\{ {{\tan }^{-1}}\left( \frac{3}{4} \right) \right\} = \frac{3}{4} \).

Final Answer

The value of \( \tan \left\{ 2{{\tan }^{-1}}\left( \frac{1}{3} \right) \right\} \) is \( \frac{3}{4} \).

Revision Table: Key Concepts for tan(2 tan⁻¹(1/3)) Calculation

Reviewing the formulas and properties used is crucial for solving such trigonometry problems.

• Formula for \( 2{{\tan }^{-1}}(x) \): \( 2{{\tan }^{-1}}(x) = {{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right) \) for \( |x| < 1 \).
• Property of inverse functions: \( \tan({{\tan }^{-1}}(y)) = y \) for all real numbers \( y \).

Additional Information: Inverse Trigonometric Formulas

Besides the formula for \( 2{{\tan }^{-1}}(x) \), there are other related formulas for inverse trigonometric functions that are useful in evaluating expressions:

• \( 2{{\tan }^{-1}}(x) = {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \) for \( |x| \le 1 \).
• \( 2{{\tan }^{-1}}(x) = {{\cos }^{-1}}\left( \frac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \) for \( x \ge 0 \).

These formulas show how \( 2{{\tan }^{-1}}(x) \) can be expressed in terms of \( {{\sin }^{-1}} \) or \( {{\cos }^{-1}} \). The choice of formula depends on the outer trigonometric function in the expression you need to evaluate.

Also recall the basic definitions:

• \( {{\sin }^{-1}}(x) \) is the angle \(\theta\) in \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) such that \( \sin(\theta) = x \).
• \( {{\cos }^{-1}}(x) \) is the angle \(\theta\) in \( [0, \pi] \) such that \( \cos(\theta) = x \).
• \( {{\tan }^{-1}}(x) \) is the angle \(\theta\) in \( (-\frac{\pi}{2}, \frac{\pi}{2}) \) such that \( \tan(\theta) = x \).