Problem 59 - A Complete Resource Book in Mathematics for JEE Main 2017

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

(A) is the correct option.

Step 1: Convert the Expression for y

We start by using the double angle identity for cosine: \( \cos 2\theta = 2\cos^2 \theta - 1 \). Thus, we can rewrite the given expression: \( y = \sqrt{\frac{1 + (2\cos^2 \theta - 1)}{1 - (2\cos^2 \theta - 1)}} = \sqrt{\frac{2\cos^2 \theta}{2 - 2\cos^2 \theta}} = \sqrt{\frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{\cos \theta}{\sin \theta} = \cot \theta \). Thus, \( y = \cot \theta \).

Step 2: Differentiate the Expression for y

We differentiate \( y = \cot \theta \) with respect to \( \theta \) using the derivative \( \frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta \). Thus, \( y' = -\csc^2 \theta \).

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

To find \( y'(\frac{\pi}{4}) \), calculate the cosecant of the angle. Since \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \), then \( \csc \frac{\pi}{4} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \). Thus, \( y'(\frac{\pi}{4}) = -\csc^2 \frac{\pi}{4} = -2 \).

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

Similarly, for \( \theta = \frac{3\pi}{4} \), \( \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \). Therefore, \( \csc \frac{3\pi}{4} = \sqrt{2} \), and \( y'(\frac{3\pi}{4}) = -\csc^2 \frac{3\pi}{4} = -2 \).

Step 5: Determine the Correct Option

Both derivatives \( y'(\frac{\pi}{4}) = -2 \) and \( y'(\frac{3\pi}{4}) = -2 \) exist and are equal, so statement A is correct. Statement B is not correct as \(-2 \cdot -2 = 4 \), not \(-4\). Statement C is incorrect because the derivatives do exist.

Key Concepts

Double Angle Identities

Double angle identities help simplify trigonometric expressions that involve angles doubled or halved. In this particular exercise, we use the double angle identity for cosine: \( \cos 2\theta = 2\cos^2 \theta - 1 \). This identity allows us to express \(\cos 2\theta\) in terms of \(\cos\theta\), making it easier to manipulate.
To convert the original expression for \(y\), we substitute \( \cos 2\theta \) in the given equation:
\[y = \sqrt{\frac{1 + (2\cos^2 \theta - 1)}{1 - (2\cos^2 \theta - 1)}}\]which simplifies to\[y = \sqrt{\frac{2\cos^2 \theta}{2 - 2\cos^2 \theta}} = \sqrt{\frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{\cos \theta}{\sin \theta} = \cot \theta \]This step reduces a complex expression into something much simpler, the cotangent of \( \theta \).

• Remember, using trigonometric identities efficiently can make solving problems much easier.
• It's vital to be familiar with these identities, as they can be crucial in various calculus and trigonometry problems.

Differentiation Techniques

Differentiation is the mathematical process of finding the derivative, or the rate of change, of a function. In this exercise, once we simplified the function to \( y = \cot \theta \), we applied differentiation to find its derivative. We used the known derivative formula for the cotangent function: \[\frac{d}{d\theta}(\cot \theta) = -\csc^2 \theta\]Differentiating \( y = \cot \theta \) with respect to \( \theta \), we obtain: \[ \ y' = -\csc^2 \theta \] The derivative \( y' \) informs us about the slope of the tangent line to the curve described by the function at any point \( \theta \).
Using differentiation techniques effectively allows you to solve problems involving rates of change, motion, and other real-world applications.

• Knowing the derivatives of basic trigonometric functions, like sine, cosine, and tangent, is fundamental to solving complex calculus problems.
• Practicing differentiation with various functions improves understanding of how functions behave and change.

Evaluating Trigonometric Expressions

Evaluating trigonometric expressions typically involve finding specific values for given angles. In our exercise, we evaluated the derivative \( y' \) at different specific angles. The steps to do this involved using known trigonometric values: - For \( \theta = \frac{\pi}{4} \): - \( \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \), thus \( \csc \frac{\pi}{4} = \sqrt{2} \) - Therefore, \( y'(\frac{\pi}{4}) = -\csc^2 \frac{\pi}{4} = -2 \)- For \( \theta = \frac{3\pi}{4} \): - \( \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} \), thus \( \csc \frac{3\pi}{4} = \sqrt{2} \) - Therefore, \( y'(\frac{3\pi}{4}) = -\csc^2 \frac{3\pi}{4} = -2 \)Both evaluations result in \(-2\), leading to a correct evaluation of statement A since both derivatives are equal. Evaluating trigonometric expressions precisely is crucial in verifying the correctness of a function's behavior at specific points.

• Ensuring you have the right trigonometric values memorized can make evaluating expressions much quicker.
• Trying out evaluations at different points tests your understanding of the function and its derivatives.