Problem 76 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Assertion: If \(\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x\), then \(\sin x=\tan ^{2} \frac{\alpha}{2}\) Reason: \(\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\)

Step-by-Step Solution

Answer

\(\sin x = 0\), consistent with \(\tan^2 \frac{\alpha}{2} = 0\).

Step 1: Understanding the Problem

We need to show that \(\sin x=\tan^{2}\frac{\alpha}{2}\) given that \(x = \cot^{-1}(\sqrt{\cos \alpha}) - \tan^{-1}(\sqrt{\cos \alpha})\). We will also use the identity \(\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)\). Our goal is to simplify the given expressions to reach the form needed for the identity of \(\sin x\).

Step 2: Apply the Difference Identity

Using the identity \(\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)\), set \(x = \sqrt{\cos \alpha}\) and \(y = \sqrt{\cos \alpha}\). Therefore, we have the expression \(\tan^{-1}\left(\frac{\sqrt{\cos \alpha} - \sqrt{\cos \alpha}}{1 + \sqrt{\cos \alpha} \cdot \sqrt{\cos \alpha}}\right)\).

Step 3: Simplify the Expression

Since the numerator is \(0\) (because \(\sqrt{\cos \alpha} - \sqrt{\cos \alpha} = 0\)), the entire expression simplifies to \(\tan^{-1}(0) = 0\). Thus, \(x = 0\).

Step 4: Find \(\sin x\)

As \(x = 0\), it follows that \(\sin x = \sin 0 = 0\).

Step 5: Connecting the Result

According to the problem, we need to show \(\sin x = \tan^2 \frac{\alpha}{2}\). By the half-angle identity, \(\tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}\). If \(\tan^2 \frac{\alpha}{2} = 0\), it implies that \(1 - \cos \alpha = 0\) or \(\cos \alpha = 1\), which is consistent as both yield \(\sin x = 0\). Therefore, the initial assertion holds under the condition \(\cos \alpha = 1\).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools in trigonometry. They allow us to find angles if we know the trigonometric value, like sine, cosine, or tangent.
In this exercise, we encounter the inverses of cotangent and tangent:

• The inverse cotangent, written as \( \cot^{-1} \), gives the angle whose cotangent is a specific number.
• The inverse tangent, \( \tan^{-1} \), provides the angle corresponding to a particular tangent value.

In the problem, we're dealing with functions subtracting one from the other: \( \cot^{-1}(\sqrt{\cos \alpha})-\tan^{-1}(\sqrt{\cos \alpha}) \).This operation isn't straightforward, so we use a special identity. Understanding these inverse trigonometric functions helps us work with angles and solve complex equations involving angles.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are always true.
They simplify expressions and solve trigonometric equations.
One key identity used in this exercise is:\[\tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)\]This identity helps manipulate subtraction between two inverse tangent functions.
By setting particular values for \( x \) and \( y \), we simplify complex expressions into manageable forms.
In this case, the identity transforms the problem, showing that if the numerator is zero, then \( \tan^{-1} \) of that expression equals zero, simplifying our work greatly.
Embracing these identities opens the door to solving many trigonometric problems efficiently.

Half-Angle Formulas

Half-angle formulas are specific identities that relate to half of a given angle in trigonometric terms.
They help express functions like cosine, sine, or tangent of an angle divided by two. A vital half-angle identity used in the solution is:\[\tan \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}}\]In our exercise, we are asked to demonstrate that \( \sin x = \tan^2 \frac{\alpha}{2} \). Combining the initial expression \( x = 0 \) with this formula, we see that \( \tan^2 \frac{\alpha}{2} = \sin x \). When \( \, x = 0, \, \sin x \) is also zero.
This is consistent under specific conditions, like when \( \cos \alpha = 1 \).
Using half-angle formulas, we transform high-level trigonometric problems into clear and manageable steps.