Problem 2 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

If \(\operatorname{cosec} x-\sin x=a^{3}, \sec x-\cos x=b^{3}\), then \(a^{2} b^{2}\) \(\left(a^{2}+b^{2}\right)\) is (a) 0 (b) 1 (c) \(-1\) (d) \(a b\).

Step-by-Step Solution

Answer

The correct answer is (a) 0

Step 1: Express Trigonometric Identities in Other Terms

Rewrite \(\operatorname{cosec} x\) as \(\frac{1}{\sin x}\) and \(\sec x\) as \(\frac{1}{\cos x}\). Thus, the identities become \(\frac{1}{\sin x} - \sin x = a^3\) and \(\frac{1}{\cos x} - \cos x = b^3\).

Step 2: Simplify the Identities

To simplify, multiply all parts of the equations by \(\sin x\) and \(\cos x\) respectively. The equations now become \(1 - \sin^2x = a^3\sin x\) and \(1 - \cos^2x = b^3\cos x\). Using Pythagorean identity which is \(\sin^2x + \cos^2x = 1\), substitute \(\sin^2 x = 1 - \cos^2 x\) in the first equation and \(\cos^2 x = 1 - \sin^2 x\) in the second. This gives the equations: \(\cos^2x = a^3\sin x\) and \(\sin^2x = b^3\cos x\) respectively

Step 3: Obtain the Expression for Required Proof

Multiply the two resulting equations from step 2. This gives \((\cos^2x)\times(\sin^2x) = a^3b^3\sin x\cos x\). Notice that \(\sin x\cos x = \frac{1}{2}\sin2x\) and \(\cos^2x * \sin^2x = \frac{1}{4}(1-\cos^2(2x))\). Substituting these identities, \(\frac{1}{4}(1-\cos^2(2x)) = \frac{1}{2}a^3b^3 \sin2x\). Simplify to get \(a^3b^3=\frac{2(1-\cos^2(2x))}{\sin2x}\). Now, squaring both sides, we get \(a^6b^6=\frac{4(1-\cos^2(2x))^2}{\sin^2(2x)}\). Using double angle formula, \(\cos(2x)=1-2\sin^2x\) and substituting in the equation, we get \(a^6b^6=\frac{4[2\sin^2x]^2}{\sin^2(2x)}\). Finally simplifying this expression gives \(a^6b^6=16\), so \(a^2b^2(a^2+b^2)=0\)

Step 4: Selecting the Correct Option

From step 3, we learn that \(a^2b^2(a^2+b^2)=0\), which matches option (a) 0. Thus, (a) 0 is the correct answer

Key Concepts

Cosecant and Secant

In trigonometry, the cosecant and secant are functions related to sine and cosine, respectively. They are reciprocals, meaning they flip the numerator and the denominator of their respective sine and cosine ratios.
For cosecant, we have:

• The function is defined as \(\operatorname{cosec} x = \frac{1}{\sin x}\).
• This is used especially when dealing with angles in right triangles.

Similarly, for secant:

• The function is defined as \(\sec x = \frac{1}{\cos x}\).
• This function is less common than cosine but equally essential in more complex trigonometric problems.

By expressing cosecant and secant in terms of sine and cosine, it becomes easier to manipulate and simplify trigonometric equations or identities. This approach, as shown in the exercise, helps simplify problems by using identities like the Pythagorean identity.

Pythagorean Identity

The Pythagorean identity is a cornerstone of trigonometry and is derived from the Pythagorean theorem. It leverages the relationship between sine and cosine:

• The fundamental identity is \(\sin^2x + \cos^2x = 1\).

This identity works for any angle \(x\) and is very useful in reducing complex trigonometric expressions.
In this particular problem, the identity helps to express one trigonometric function in terms of another. For example, \(\sin^2x = 1 - \cos^2x\) and \(\cos^2x = 1 - \sin^2x\).
By substituting these relationships into the equation, it becomes easier to solve and verify trigonometric equations as demonstrated in the step-by-step solution.

Double Angle Formulas

Double angle formulas are special trigonometric identities that express trigonometric functions of double angles, such as \(2x\), in terms of single angle values.

• The formula for cosine is \(\cos(2x) = 1 - 2\sin^2x = 2\cos^2x - 1\).
• The formula for sine is \(\sin(2x) = 2\sin x \cos x\).

These formulas are particularly helpful in the exercise, as you can use them to simplify expressions involving trigonometric functions squared or multiplied. For instance, recognizing that \(\sin x \cos x = \frac{1}{2}\sin 2x\) aids in transforming expressions like those found in the solution steps.
Furthermore, employing these formulas enables the simplification of complex expressions involving products or ratios of trigonometric functions, making the problem-solving process more efficient and straightforward.