Problem 80 - A Complete Resource Book in Mathematics for JEE Main

Question

10\. \(\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=\mathrm{x}\), then \(\sin \mathrm{x}\) is equal to: (A)tan \(^{2}\left(\frac{\alpha}{2}\right)\) (B) \(\cot ^{2}\left(\frac{\alpha}{2}\right)\) (C) \(\tan \alpha\) (D) \(\cot \left(\frac{\alpha}{2}\right)\)

Step-by-Step Solution

Answer

(A) \(\tan^2\left(\frac{\alpha}{2}\right)\)

Step 1: Use the identity cot⁻¹(t) + tan⁻¹(t) = π/2

We know that \(\cot^{-1}(t) + \tan^{-1}(t) = \frac{\pi}{2}\) for all \(t > 0\).
Let \(t = \sqrt{\cos \alpha}\). Then \(\cot^{-1}(t) = \frac{\pi}{2} - \tan^{-1}(t)\).

Step 2: Express x in terms of tan⁻¹

\(x = \cot^{-1}(\sqrt{\cos \alpha}) - \tan^{-1}(\sqrt{\cos \alpha}) = \frac{\pi}{2} - 2\tan^{-1}(\sqrt{\cos \alpha})\)

Step 3: Find sin x

Let \(\theta = \tan^{-1}(\sqrt{\cos \alpha})\), so \(\tan \theta = \sqrt{\cos \alpha}\).
Then \(x = \frac{\pi}{2} - 2\theta\), so \(\sin x = \sin\left(\frac{\pi}{2} - 2\theta\right) = \cos 2\theta\).

Step 4: Compute cos 2θ

Using \(\cos 2\theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1 - \cos \alpha}{1 + \cos \alpha}\).
Using the half-angle identities: \(1 - \cos \alpha = 2\sin^2\frac{\alpha}{2}\) and \(1 + \cos \alpha = 2\cos^2\frac{\alpha}{2}\).
Therefore \(\sin x = \frac{2\sin^2\frac{\alpha}{2}}{2\cos^2\frac{\alpha}{2}} = \tan^2\left(\frac{\alpha}{2}\right)\).
The answer is (A) \(\tan^2\left(\frac{\alpha}{2}\right)\).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental equations used in trigonometry that relate the trigonometric functions of an angle to one another. They help simplify complex trigonometric expressions and prove equivalence between different conditions. One of the most common identities is the Pythagorean identity, which states \[ \\sin^2\theta + \cos^2\theta = 1. \]It is especially useful when dealing with transformations between sine and cosine. Another vital class of identities is angle sum and difference identities, which provide relationships for trigonometric functions of angles like \( \sin(A \pm B) \) and \( \cos(A \pm B) \). These identities are indispensable in solving a wide range of problems that involve adding and subtracting angles.Moreover, reciprocal identities relate trigonometric functions with their respective reciprocal functions, such as \( \cot\theta = \frac{1}{\tan\theta} \) and \( \tan\theta = \frac{1}{\cot\theta} \). They allow for alternative ways to express and solve equations involving these functions. Understanding how to apply these identities effectively is key to mastering trigonometry.

Cotangent Function

The cotangent function is one of the six fundamental trigonometric functions. It is the reciprocal of the tangent function, meaning \( \cot \theta = \frac{1}{\tan \theta} \). Essentially, if you know the tangent of an angle, you can easily find its cotangent by taking the reciprocal. In right triangle terms, the cotangent of an angle is the ratio of the length of the adjacent side to the length of the opposite side.When working with inverse trigonometric functions, such as \( \text{cot}^{-1}(x) \), we retrieve the original angle that corresponds to the given cotangent value. Understanding this relationship is crucial, especially when dealing with expressions that involve transformations and comparisons between cotangent and other trigonometric functions. For example, in the execution of complex trigonometric equations or the conversion between different forms, knowing that \( \cot \beta = \sqrt{ \cos \alpha} \) allows you to interact seamlessly with different parts of the problem, leveraging the known cosine value to ascertain more about your angles.

Tangent Function

The tangent function is another key player among the trigonometric functions, denoted by \( \tan \theta \). It is defined as the ratio of the sine and cosine of an angle, specifically \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). In practical scenarios involving right triangles, it represents the ratio of the length of the opposite side to the adjacent side.Understanding the \( \tan \theta \) is imperative, particularly when solving equations that involve its inverse function \( \text{tan}^{-1}(x) \), used to find angles from given tangent values. As in the presented exercise, where working with both tangent and its inverse allows simplification and understanding of the problem space. The tangent function is periodic with a period of \( \pi \), making it useful in scenarios involving periodic phenomena, like waves or vibrations. Mastering how the tangent relates to and interacts with other trigonometric functions is vital for interpreting more complex mathematical scenarios, whether within pure mathematics or applied contexts.