Analyzing Inverse Trigonometric Identities

We are asked to evaluate the correctness of two given mathematical statements involving inverse trigonometric functions. Let's analyze each statement individually.

Statement 1: \({\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{3}{5} = \frac{\pi }{2}\)

To check this statement, we can use a property of inverse trigonometric functions. Recall the identity: \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\), for \(-1 \le x \le 1\).

Let's try to convert one of the terms, say \({\sin ^{ - 1}}\frac{3}{5}\), into a cosine inverse term. Let \(\theta = {\sin ^{ - 1}}\frac{3}{5}\). This means \(\sin\theta = \frac{3}{5}\).

Consider a right-angled triangle where the opposite side is 3 and the hypotenuse is 5. By the Pythagorean theorem, the adjacent side would be \(\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\).

Now, we can find the cosine of the angle \(\theta\). \(\cos\theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{4}{5}\). Therefore, \(\theta = {\cos ^{ - 1}}\frac{4}{5}\).

So, we can rewrite the original statement as:

\({\sin ^{ - 1}}\frac{4}{5} + {\cos ^{ - 1}}\frac{4}{5} = \frac{\pi }{2}\)

This matches the identity \(\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\) with \(x = \frac{4}{5}\). Since \(x = \frac{4}{5}\) is between -1 and 1, the identity holds true.

Alternatively, we could use the identity \(\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})\) for \(x \ge 0, y \ge 0\) and \(x^2 + y^2 \le 1\). Here \(x=4/5\) and \(y=3/5\). \(x \ge 0, y \ge 0\) is true. \(x^2 + y^2 = (4/5)^2 + (3/5)^2 = 16/25 + 9/25 = 25/25 = 1\). Since \(x^2+y^2=1\), the identity simplifies to \(\sin^{-1}(1) = \pi/2\).

\(\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{3}{5} = \sin^{-1}\left(\frac{4}{5}\sqrt{1 - \left(\frac{3}{5}\right)^2} + \frac{3}{5}\sqrt{1 - \left(\frac{4}{5}\right)^2}\right)\)

\(= \sin^{-1}\left(\frac{4}{5}\sqrt{1 - \frac{9}{25}} + \frac{3}{5}\sqrt{1 - \frac{16}{25}}\right)\)

\(= \sin^{-1}\left(\frac{4}{5}\sqrt{\frac{16}{25}} + \frac{3}{5}\sqrt{\frac{9}{25}}\right)\)

\(= \sin^{-1}\left(\frac{4}{5} \cdot \frac{4}{5} + \frac{3}{5} \cdot \frac{3}{5}\right)\)

\(= \sin^{-1}\left(\frac{16}{25} + \frac{9}{25}\right)\)

\(= \sin^{-1}\left(\frac{25}{25}\right)\)

\(= \sin^{-1}(1)\)

Since \(\sin(\pi/2) = 1\) and the range of \(\sin^{-1}x\) is \([-\pi/2, \pi/2]\), \(\sin^{-1}(1) = \pi/2\).

Thus, the statement \({\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{3}{5} = \frac{\pi }{2}\) is correct.

Statement 2: \({\tan ^{ - 1}}\sqrt 3 + {\tan ^{ - 1}}1 = - {\tan ^{ - 1}}\left( {2 + \sqrt 3 } \right)\)

Let's evaluate the left-hand side (LHS) of the statement:

LHS = \({\tan ^{ - 1}}\sqrt 3 + {\tan ^{ - 1}}1\)

We know that \(\tan(\pi/3) = \sqrt{3}\), so \({\tan ^{ - 1}}\sqrt 3 = \pi/3\). The principal value \(\pi/3\) is in the range \((-\pi/2, \pi/2)\).

We also know that \(\tan(\pi/4) = 1\), so \({\tan ^{ - 1}}1 = \pi/4\). The principal value \(\pi/4\) is in the range \((-\pi/2, \pi/2)\).

LHS = \(\frac{\pi}{3} + \frac{\pi}{4}\)

To add these fractions, find a common denominator, which is 12:

LHS = \(\frac{4\pi}{12} + \frac{3\pi}{12} = \frac{7\pi}{12}\)

Now let's evaluate the right-hand side (RHS) of the statement:

RHS = \(- {\tan ^{ - 1}}\left( {2 + \sqrt 3 } \right)\)

We need to find the value of \({\tan ^{ - 1}}\left( {2 + \sqrt 3 } \right)\). We know that \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\). Let's consider \(\tan(75^\circ)\) or \(\tan(5\pi/12)\), which can be written as \(\tan(45^\circ + 30^\circ)\) or \(\tan(\pi/4 + \pi/6)\).

\(\tan\left(\frac{\pi}{4} + \frac{\pi}{6}\right) = \frac{\tan(\pi/4) + \tan(\pi/6)}{1 - \tan(\pi/4) \tan(\pi/6)} = \frac{1 + 1/\sqrt{3}}{1 - 1 \cdot 1/\sqrt{3}} = \frac{(\sqrt{3}+1)/\sqrt{3}}{(\sqrt{3}-1)/\sqrt{3}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}\)

Rationalize the denominator:

\(\frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{(\sqrt{3})^2 + 1^2 + 2\sqrt{3}}{(\sqrt{3})^2 - 1^2} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}\)

So, \(\tan(5\pi/12) = 2 + \sqrt{3}\).

Since \(5\pi/12\) is in the range \((-\pi/2, \pi/2)\) of \({\tan ^{ - 1}}x\), we have \({\tan ^{ - 1}}\left( {2 + \sqrt 3 } \right) = \frac{5\pi}{12}\).

Now substitute this back into the RHS:

RHS = \(-\frac{5\pi}{12}\)

Comparing the LHS and RHS:

LHS = \(\frac{7\pi}{12}\)

RHS = \(-\frac{5\pi}{12}\)

\(\frac{7\pi}{12} \neq -\frac{5\pi}{12}\).

Thus, the statement \({\tan ^{ - 1}}\sqrt 3 + {\tan ^{ - 1}}1 = - {\tan ^{ - 1}}\left( {2 + \sqrt 3 } \right)\) is incorrect.

Summary of Findings

• Statement 1: \({\sin ^{ - 1}}\frac{4}{5} + {\sin ^{ - 1}}\frac{3}{5} = \frac{\pi }{2}\) is Correct.
• Statement 2: \({\tan ^{ - 1}}\sqrt 3 + {\tan ^{ - 1}}1 = - {\tan ^{ - 1}}\left( {2 + \sqrt 3 } \right)\) is Incorrect.

Based on our analysis, only statement 1 is correct.

Revision Table: Inverse Trigonometric Identities

Useful Identities:

• \(\sin^{-1}x + \cos^{-1}x = \pi/2\), for \(-1 \le x \le 1\)
• \(\tan^{-1}x + \cot^{-1}x = \pi/2\), for \(-\infty < x < \infty\)
• \(\sec^{-1}x + \csc^{-1}x = \pi/2\), for \(|x| \ge 1\)
• \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)\), if \(xy < 1\)
• \(\tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)\), if \(x > 0, y > 0, xy > 1\)
• \(\tan^{-1}x + \tan^{-1}y = -\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)\), if \(x < 0, y < 0, xy > 1\)

Additional Information on Inverse Trigonometric Functions

Inverse trigonometric functions, also known as arc functions, are the inverse functions of the trigonometric functions (sine, cosine, tangent, etc.). They are used to find the angle when the value of the trigonometric ratio is known. Because trigonometric functions are periodic, their inverse functions are multi-valued. To make them single-valued, we restrict their domains, which leads to the concept of principal values and principal ranges.

Understanding the principal range of each inverse trigonometric function is crucial for correctly evaluating expressions and using identities. For example, the range of \(\sin^{-1}x\) is \([-\pi/2, \pi/2]\), meaning the output of \(\sin^{-1}x\) will always be an angle within this interval. Similarly, the range of \(\tan^{-1}x\) is \((-\pi/2, \pi/2)\).

When dealing with identities involving sums or differences of inverse tangent functions, it's important to consider the product \(xy\) to determine which form of the identity is applicable. As shown in Statement 2, ignoring this condition can lead to incorrect results.