Problem 59 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(y=\sqrt{\frac{1+\cos 2 \theta}{1-\cos 2 \theta}}\), then (A) \(y^{\prime}\left(\frac{\pi}{4}\right)=y^{\prime}\left(\frac{3 \pi}{4}\right)\) (B) \(y^{\prime}\left(\frac{\pi}{4}\right) \cdot y^{\prime}\left(\frac{3 \pi}{4}\right)=-4\) (C) \(y^{\prime}\left(\frac{\pi}{4}\right)\) and \(y^{\prime}\left(\frac{3 \pi}{4}\right)\) do not exist (D) None of these

Step-by-Step Solution

Answer

Option (A) is correct: \(y'\left(\frac{\pi}{4}\right) = y'\left(\frac{3\pi}{4}\right)\).

Step 1: Rewrite the Expression

First, recall and use the half-angle identities. The identity \( \cos 2\theta = 1 - 2\sin^2 \theta \) can be rearranged to express \(1 + \cos 2\theta\) and \(1 - \cos 2\theta\) in terms of \(\sin\theta\). Use the identities:\[1 + \cos 2\theta = 2\cos^2 \theta\]\[1 - \cos 2\theta = 2\sin^2 \theta\]Thus, the expression becomes:\[y = \sqrt{\frac{2\cos^2 \theta}{2\sin^2 \theta}} = \sqrt{\frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{\cos \theta}{\sin \theta} = \cot \theta\]

Step 2: Differentiate the Function

The function is now \(y = \cot \theta\). The derivative of \(\cot \theta\) with respect to \(\theta\) is \(-\csc^2 \theta\). Therefore, \[y' = -\csc^2 \theta\]

Step 3: Evaluate Derivative at \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\)

Using the derivative derived in Step 2, evaluate:- For \(\theta = \frac{\pi}{4}\), \( y'\left(\frac{\pi}{4}\right) = -\csc^2 \left(\frac{\pi}{4}\right) = -2\).- For \(\theta = \frac{3\pi}{4}\), \( y'\left(\frac{3\pi}{4}\right) = -\csc^2 \left(\frac{3\pi}{4}\right) = -2\).

Step 4: Analyze the Statements

Now examine the given options:- (A) checks if \(y'\left(\frac{\pi}{4}\right) = y'\left(\frac{3\pi}{4}\right)\). Since both are -2, this is true.- (B) involves the product \(y'\left(\frac{\pi}{4}\right) \cdot y'\left(\frac{3\pi}{4}\right) = (-2) \cdot (-2) = 4\), which is not -4.- (C) suggests the derivatives do not exist, which is false since both exist and equal -2.- Given the above, only (A) is correct.

Key Concepts

Half-Angle Identities

In trigonometry, half-angle identities are useful for simplifying expressions that involve trigonometric functions of half-angles. For example, when dealing with trigonometric functions like cosine, one often uses the identity: \[ \cos(2\theta) = 1 - 2\sin^2(\theta) \] This identity is derived from the double angle formulas and helps to rewrite expressions. For instance, the expressions \(1 + \cos 2\theta\) and \(1 - \cos 2\theta\) can be rewritten using the half-angle identities:- \(1 + \cos 2\theta = 2\cos^2(\theta)\) - \(1 - \cos 2\theta = 2\sin^2(\theta)\) These transformations are particularly useful when trying to simplify complex algebraic fractions that involve trigonometric functions. In this specific problem, these identities allow the expression under the square root to be simplified, transforming an intricate square root expression into a simpler form, such as \( \cot \theta \). Understanding and applying half-angle identities can greatly simplify calculus problems, particularly when an equation looks complex.

Cotangent Function

The cotangent function, often denoted as \( \cot \theta \), is the reciprocal of the tangent function. It is defined as:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \] This function is critical in trigonometry because it provides an alternative way of looking at angles and can be quite handy when simplifying equations involving right triangles.
In our exercise, after applying half-angle identities, the equation simplifies to \( \cot \theta \). Recognizing that a complicated equation reduced to this function shows the power of trigonometric identities in simplifying expressions.

• Cotangent is undefined at angles where \( \sin \theta = 0 \), for instance, multiples of \( \pi \).
• Many trigonometric identities involve \( \cot \theta \), making familiarity with this function useful for solving trig-related problems.

When working with derivatives, knowing the expression and applications of \( \cot \theta \) can simplify expressions significantly.

Differentiation

In calculus, differentiation is the process of finding the derivative, or the rate at which a function changes. The derivative of a function provides information about its slope or gradient at any given point. Differentiation is a fundamental concept because it helps to analyze changes in functions. In trigonometry, differentiating trigonometric functions is a key skill.
For the cotangent function specifically, the derivative with respect to \( \theta \) is:\[ \frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta \] Here, \( \csc \theta \) is the cosecant function, which is the reciprocal of the sine function.
In our exercise, after simplifying the function, we use the derivative of \( \cot \theta \) to find the rates of change at different points:

• \( y'(\theta) = -\csc^2(\theta) \)
• This negative value indicates that the function has a decreasing slope at every point where \( \csc^2(\theta) \) is defined.

Evaluation at \( \theta = \frac{\pi}{4} \) and \( \theta = \frac{3\pi}{4} \) then shows both derivatives to be \(-2\), demonstrating a consistent rate of decrease at those specific angles.