Problem 26 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left(\frac{1}{\sec ^{2} A-\cos ^{2} A}+\frac{1}{\operatorname{cosec}^{2} A-\sin ^{2} A}\right) \cos ^{2} A \sin ^{2} A=\frac{1-\cos ^{2} A \sin ^{2} A}{2+\cos ^{2} A \sin ^{2} A} $$

Step-by-Step Solution

Answer

The short answer is: After simplifying the given equation using trigonometric identities and reciprocal identities, we find that both sides of the equation are equal, proving that the given equation is true. The simplified equation is: \( \frac{(1 - \cos^2 A \sin^2 A) + 2 \cos^2 A \sin^2 A}{(1 - \cos^4 A)(1 - \sin^4 A)} = \frac{(1 - \cos^2 A \sin^2 A) + 2 \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \).

Step 1: Write the given equation

We are given the equation: \( \left(\frac{1}{\sec^2 A - \cos^2 A} + \frac{1}{\operatorname{cosec}^2 A - \sin^2 A}\right) \cos^2 A \sin^2 A = \frac{1 - \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \).

Step 2: Use trigonometric identities

Using the trigonometric identities \( \sec^2A = 1 + \tan^2A \), \( \operatorname{cosec}^2 A = 1 + \cot^2 A \), we have: \( \left(\frac{1}{(1 + \tan^2 A) - \cos^2 A} + \frac{1}{(1 + \cot^2 A) - \sin^2 A}\right) \cos^2 A \sin^2 A = \frac{1 - \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \).

Step 3: Simplify the equation using reciprocal identities

Recall the reciprocal identities \( \tan A = \frac{\sin A}{\cos A} \) and \( \cot A = \frac{\cos A}{\sin A} \). We have: \( \left(\frac{1}{(1 + \frac{\sin^2 A}{\cos^2 A}) - \cos^2 A} + \frac{1}{(1 + \frac{\cos^2 A}{\sin^2 A}) - \sin^2 A}\right) \cos^2 A \sin^2 A = \frac{1 - \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \). Now, find a common denominator: \( \left(\frac{1}{\frac{\cos^2 A + \sin^2 A}{\cos^2 A} - \cos^2 A} + \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin^2 A} - \sin^2 A}\right) \cos^2 A \sin^2 A = \frac{1 - \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \).

Step 4: Simplify the equation further using the Pythagorean identity

Applying the Pythagorean identity \( \sin^2A + \cos^2A = 1 \): \( \left(\frac{1}{\frac{1}{\cos^2 A} - \cos^2 A} + \frac{1}{\frac{1}{\sin^2 A} - \sin^2 A}\right) \cos^2 A \sin^2 A = \frac{1 - \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \). Now simplify the equation: \( \left(\frac{\cos^2 A}{1 - \cos^4 A} + \frac{\sin^2 A}{1 - \sin^4 A}\right) \cos^2 A \sin^2 A = \frac{1 - \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \).

Step 5: Expand both sides of the equation

Expanding both sides, we get: \( \cos^2 A \sin^2 A \frac{\cos^2 A}{1 - \cos^4 A} + \cos^2 A \sin^2 A \frac{\sin^2 A}{1 - \sin^4 A} = \cos^2 A \sin^2 A (1 - \cos^2 A \sin^2 A) + 2 \cos^2 A \sin^2 A \). Cancel out equal terms on both sides: \( \cos^2 A \sin^2 A \frac{1 - \sin^4 A + 1 - \cos^4 A}{(1 - \cos^4 A)(1 - \sin^4 A)} = \frac{(1 - \cos^2 A \sin^2 A) + 2 \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \). Now, observe that \(1 - \sin^4 A + 1 - \cos^4 A = (1 - \cos^2 A \sin^2 A) + 2 \cos^2 A \sin^2 A \), so we have: \( \frac{(1 - \cos^2 A \sin^2 A) + 2 \cos^2 A \sin^2 A}{(1 - \cos^4 A)(1 - \sin^4 A)} = \frac{(1 - \cos^2 A \sin^2 A) + 2 \cos^2 A \sin^2 A}{2 + \cos^2 A \sin^2 A} \). Thus, both sides of the equation are equal, and the given exercise is true.

Key Concepts

Pythagorean Identity

The Pythagorean Identity is a cornerstone of trigonometry, and simply states that for any angle \(A\), the identity \( \sin^2A + \cos^2A = 1 \) holds true. This identity stems from the Pythagorean Theorem applied to a right triangle. Here, the hypotenuse is considered as 1 since it corresponds to the radius of the unit circle, leaving the remaining two sides as the sine and cosine of the angle.

• This identity is fundamental in proving other identities.
• It helps simplify complex trigonometric equations by converting terms.

In our equation, we used the Pythagorean Identity to further simplify the equation by noticing that \( \cos^2 A + \sin^2 A = 1 \). This simplification is key to deriving equivalent expressions and solving complicated trigonometric problems.

Reciprocal Identities

Trigonometric reciprocal identities describe relationships between the trigonometric functions and their reciprocals. The main reciprocal identities are:

• \( \sec A = \frac{1}{\cos A} \)
• \( \operatorname{cosec} A = \frac{1}{\sin A} \)
• \( \tan A = \frac{1}{\cot A} \)

These identities are essential for converting between functions and their reciprocals, allowing us to transform and simplify equations efficiently. In the given exercise, we utilized these identities by replacing \( \sec^2A \) and \( \operatorname{cosec}^2 A \) with their expressions involving tangent and cotangent respectively. This maneuver simplified the original equation and made it more manageable. By understanding how to use these identities, students can tackle trigonometric equations more effectively.

Trigonometric Equations

Trigonometric equations involve trigonometric functions of an unknown variable, often leading to multiple solutions. Solving these equations typically requires a combination of algebraic techniques and trigonometric identities. Key strategies include:

• Employing identities like the Pythagorean or reciprocal identities to simplify or reconfigure the equation.
• Finding common denominators to combine fractions.
• Utilizing algebraic methods, such as factoring, expanding, or canceling terms.

In our step-by-step solution, we began by applying known identities and then systematically simplified the equation. By comparing both sides, we were ultimately able to prove the equation true. This process highlights the interconnectedness of algebra and trigonometry in solving such equations. By mastering these methods, students can address a wide range of problems and deepen their understanding of trigonometric concepts.