Evaluating Statements on Inverse Trigonometric Functions

The question asks us to determine the correctness of two given statements involving inverse trigonometric functions. Let's analyze each statement individually.

Analyzing Statement 1: Inverse Tangent Property

Statement 1 says: There exists \({\rm{\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) for which tan^-1(tan \({\rm{\theta }}\)) \(\ne\) \({\rm{\theta }}\).

Let's understand the property of tan^-1(tan \({\rm{\theta }}\)).

• The domain of the function tan \({\rm{\theta }}\) is all real numbers except \({\rm{\theta }} = n{\rm{\pi }} + \frac{{\rm{\pi }}}{2}\), where \(n\) is an integer.
• The principal value branch of tan^-1(\(x\)) has a range of \(\left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\).

The identity tan^-1(tan \({\rm{\theta }}\)) = \({\rm{\theta }}\) holds true if and only if \({\rm{\theta }}\) lies within the principal value branch of tan^-1, which is the interval \(\left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\). If \({\rm{\theta }}\) is outside this interval, tan^-1(tan \({\rm{\theta }}\)) will be equal to the value in \(\left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) that has the same tangent as \({\rm{\theta }}\).

Statement 1 claims that there exists a \({\rm{\theta }}\) in the interval \(\left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) such that tan^-1(tan \({\rm{\theta }}\)) \(\ne\) \({\rm{\theta }}\). However, by the definition and property of the principal value branch, for any \({\rm{\theta }}\) in \(\left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\), tan^-1(tan \({\rm{\theta }}\)) is always equal to \({\rm{\theta }}\).

Therefore, Statement 1 is incorrect.

Analyzing Statement 2: Difference of Inverse Sines

Statement 2 says: \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{15}}} \right)\)

We will use the formula for the difference of inverse sines: \({\sin ^{ - 1}}(x) - {\sin ^{ - 1}}(y) = {\sin ^{ - 1}}(x\sqrt{1-y^2} - y\sqrt{1-x^2})\), provided that \(x, y \in [-1, 1]\) and \(x^2 + y^2 \le 1\) or \(xy \le 0\).

In this statement, we have \(x = \frac{1}{3}\) and \(y = \frac{1}{5}\). Both \(x\) and \(y\) are in the interval \([-1, 1]\). Let's check the condition \(x^2 + y^2 \le 1\): \(x^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\) \(y^2 = \left(\frac{1}{5}\right)^2 = \frac{1}{25}\) \(x^2 + y^2 = \frac{1}{9} + \frac{1}{25} = \frac{25 + 9}{225} = \frac{34}{225}\). Since \(\frac{34}{225} < 1\), the condition \(x^2 + y^2 \le 1\) is satisfied. We can use the formula directly.

Left Hand Side (LHS): \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right)\)

Applying the formula:

\({\sin ^{ - 1}}\left(\frac{1}{3}\sqrt{1-\left(\frac{1}{5}\right)^2} - \frac{1}{5}\sqrt{1-\left(\frac{1}{3}\right)^2}\right)\)

\(= {\sin ^{ - 1}}\left(\frac{1}{3}\sqrt{1-\frac{1}{25}} - \frac{1}{5}\sqrt{1-\frac{1}{9}}\right)\)

\(= {\sin ^{ - 1}}\left(\frac{1}{3}\sqrt{\frac{25-1}{25}} - \frac{1}{5}\sqrt{\frac{9-1}{9}}\right)\)

\(= {\sin ^{ - 1}}\left(\frac{1}{3}\sqrt{\frac{24}{25}} - \frac{1}{5}\sqrt{\frac{8}{9}}\right)\)

\(= {\sin ^{ - 1}}\left(\frac{1}{3} \cdot \frac{\sqrt{24}}{5} - \frac{1}{5} \cdot \frac{\sqrt{8}}{3}\right)\)

\(= {\sin ^{ - 1}}\left(\frac{\sqrt{24}}{15} - \frac{\sqrt{8}}{15}\right)\)

We can simplify the square roots:

\(\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}\)

\(\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}\)

Substituting these back:

\(= {\sin ^{ - 1}}\left(\frac{2\sqrt{6}}{15} - \frac{2\sqrt{2}}{15}\right)\)

\(= {\sin ^{ - 1}}\left(\frac{2\sqrt{6} - 2\sqrt{2}}{15}\right)\)

\(= {\sin ^{ - 1}}\left(\frac{2(\sqrt{6} - \sqrt{2})}{15}\right)\)

Now, let's manipulate the term inside the sin^-1 to match the RHS of Statement 2:

\(\frac{2(\sqrt{6} - \sqrt{2})}{15} = \frac{2\sqrt{2}(\sqrt{3} - 1)}{15}\)

This matches the term on the Right Hand Side (RHS) of Statement 2.

So, LHS = RHS. Therefore, Statement 2 is correct.

Conclusion

Based on the analysis:

• Statement 1 is incorrect because tan^-1(tan \({\rm{\theta }}\)) = \({\rm{\theta }}\) for all \({\rm{\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\).
• Statement 2 is correct as verified by the inverse sine difference formula.

Thus, only Statement 2 is correct.

Revision Table: Key Inverse Trigonometric Properties

Additional Information: Formulas for Inverse Sine

Here are some useful formulas involving the inverse sine function:

• Sum: \({\sin ^{ - 1}}(x) + {\sin ^{ - 1}}(y) = {\sin ^{ - 1}}(x\sqrt{1-y^2} + y\sqrt{1-x^2})\), applicable under certain conditions for \(x, y \in [-1, 1]\). For example, if \(x \ge 0, y \ge 0\) and \(x^2 + y^2 \le 1\), or if \(xy < 0\) and \(x^2+y^2 \ge 1\).
• Difference: \({\sin ^{ - 1}}(x) - {\sin ^{ - 1}}(y) = {\sin ^{ - 1}}(x\sqrt{1-y^2} - y\sqrt{1-x^2})\), applicable under certain conditions for \(x, y \in [-1, 1]\). For example, if \(x \ge 0, y \ge 0\) and \(x^2 + y^2 \le 1\).
• Note that the formulas for sum and difference have variations depending on the signs of \(x\) and \(y\) and the value of \(x^2+y^2\) to ensure the result falls within the principal value branch \(\left[ - \frac{{\rm{\pi}}}{2}, \frac{{\rm{\pi}}}{2} \right]\). However, for \(x, y \in [0, 1]\) and \(x^2+y^2 \le 1\), the basic formulas without adjustments apply. In Statement 2, \(x=1/3\) and \(y=1/5\) are in \([0, 1]\) and \(x^2+y^2 = 34/225 \le 1\), so the simple formula was used.