Problem 5 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

A value of \(\tan ^{-1}\left(\sin \left(\cos ^{-1}\left(\sqrt{\frac{2}{3}}\right)\right)\right)\) is (a) \(\frac{\pi}{4}\) (b) \(\frac{\pi}{2}\) (c) \(\frac{\pi}{3}\) (d) \(\frac{\pi}{6}\)

Step-by-Step Solution

Answer

The value is \( \frac{\pi}{6} \). The correct option is (d).

Step 1: Understanding the Inner Function

Start with the expression \( \cos^{-1}\left(\sqrt{\frac{2}{3}}\right) \). This represents the angle \( \theta \) such that \( \cos \theta = \sqrt{\frac{2}{3}} \). This angle is in the range \( [0, \pi] \) since \( \cos^{-1} \) outputs angles in this range.

Step 2: Finding \( \sin(\theta) \)

Since \( \theta = \cos^{-1}\left(\sqrt{\frac{2}{3}}\right) \), we use the Pythagorean identity: \( \sin^2 \theta = 1 - \cos^2 \theta \). Thus, \( \sin^2 \theta = 1 - \left(\sqrt{\frac{2}{3}}\right)^2 = 1 - \frac{2}{3} = \frac{1}{3} \). Therefore, \( \sin \theta = \sqrt{\frac{1}{3}} = \frac{1}{\sqrt{3}} \).

Step 3: Handling the \( \tan^{-1} \) Function

Now consider \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \). Recall that \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \). Therefore, \( \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \).

Step 4: Choose the Correct Option

We found that \( \tan^{-1}\left(\sin \left(\cos^{-1} \left(\sqrt{\frac{2}{3}}\right)\right)\right) = \frac{\pi}{6} \). Therefore, the correct option is (d) \( \frac{\pi}{6} \).

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental concept in trigonometry that connects the squares of sine and cosine functions. Mathematically, it is expressed as: \[\sin^2 \theta + \cos^2 \theta = 1\]This identity is derived from the Pythagorean Theorem, which relates to triangles. It states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. In the unit circle, this translates to the relationship between the sine and cosine of an angle.When we are given \(\theta = \cos^{-1}(\sqrt{\frac{2}{3}})\), we know that \(\cos \theta = \sqrt{\frac{2}{3}}\). We can use the Pythagorean identity to find \(\sin \theta\) by rearranging the equation to find:\[\sin^2 \theta = 1 - \cos^2 \theta\]Substituting \(\cos\theta\), we calculate \(\sin^2 \theta = 1 - \left(\sqrt{\frac{2}{3}}\right)^2 = \frac{1}{3}\). Therefore, \(\sin \theta = \frac{1}{\sqrt{3}}\), which is crucial in solving inverse trigonometric problems like the one in the exercise.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They are essential tools in solving trigonometric equations and simplifying expressions. Some key trigonometric identities include:

• Pythagorean Identity: \(\sin^2 \theta + \cos^2 \theta = 1\)
• Quotient Identities: \(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
• Reciprocal Identities: \(\csc \theta = \frac{1}{\sin \theta}, \sec \theta = \frac{1}{\cos \theta}, \cot \theta = \frac{1}{\tan \theta}\)

In the exercise, we primarily utilized the Pythagorean identity to evaluate inverse trigonometric functions. By simplifying \(\sin(\theta)\) using the identity, we were able to determine \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\). This evaluates to \(\frac{\pi}{6}\) because it directly corresponds to a well-known angle of 30 degrees, or \(\pi/6\), whose tangent value is \(\frac{1}{\sqrt{3}}\). Understanding these identities helps in transforming complex trigonometric expressions into more manageable forms.

Angle Evaluation

Angle evaluation is a vital step in solving trigonometric functions, especially inverse trigonometric functions. The goal is often to find an angle whose trigonometric function equals a known value. Consider the expression \(\tan^{-1}\left(\sin \left(\cos^{-1} \left(\sqrt{\frac{2}{3}}\right)\right)\right)\). Here, \(\cos^{-1}(\sqrt{\frac{2}{3}})\) led us to determine angle \(\theta\). The angle \(\theta\) has a cosine value of \(\sqrt{\frac{2}{3}}\). To evaluate such expressions:

• Find the value of the trigonometric function (like sine or cosine) using known identities.
• Identify the angle associated with this value (often by recognizing common angle values from trigonometric tables or unit circle knowledge).

In this context, by realizing \(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\) corresponds to \(\frac{\pi}{6}\), the angle evaluation was concluded. Understanding typical angle measures and their trigonometric values is critical for quickly solving such problems. This clarity enables the determination of angles for inverse trigonometric functions with ease.