Class 12 RS Aggarwal Chapter-4 Inverse Trigonometric Functions Solutions - Free PDF Download
FAQs on RS Aggarwal Class 12 Solutions Chapter-4 Inverse Trigonometric Functions
1. How do I find the principal value of an inverse trigonometric function like sin⁻¹(-1/2) as per the CBSE 2025-26 syllabus?
To find the principal value, you must use the specific range defined for each inverse trigonometric function. For sin⁻¹(x), the principal value branch is [-π/2, π/2]. The steps are:
- First, let y = sin⁻¹(-1/2). This means sin(y) = -1/2.
- We know that sin(π/6) = 1/2. Using the property sin(-θ) = -sin(θ), we get sin(-π/6) = -1/2.
- The value -π/6 lies within the principal value range of [-π/2, π/2].
- Therefore, the principal value of sin⁻¹(-1/2) is -π/6.
2. What is the step-by-step method for simplifying complex expressions like tan⁻¹(cos x / (1 - sin x)) found in RS Aggarwal solutions?
The core strategy is to convert the inner expression into a form of tan(θ) so you can use the property tan⁻¹(tan θ) = θ. The steps are as follows:
- Use trigonometric identities to simplify the expression inside the bracket. For cos x / (1 - sin x), you can use cos x = cos²(x/2) - sin²(x/2) and 1 = sin²(x/2) + cos²(x/2), and sin x = 2sin(x/2)cos(x/2).
- Substitute these into the expression. It simplifies to (cos(x/2) + sin(x/2)) / (cos(x/2) - sin(x/2)).
- Divide the numerator and denominator by cos(x/2) to get it in terms of tan. The expression becomes (1 + tan(x/2)) / (1 - tan(x/2)).
- This is the formula for tan(π/4 + x/2).
- So, the original expression becomes tan⁻¹(tan(π/4 + x/2)), which simplifies to π/4 + x/2.
3. How are properties like sin⁻¹(x) + cos⁻¹(x) = π/2 used to solve problems in RS Aggarwal Class 12 Chapter 4?
This property is a powerful tool for simplification and solving equations. In the RS Aggarwal solutions, it is often used to:
- Solve Equations: If you have an equation like sin⁻¹(x) + cos⁻¹(1/2) = π/3, you can use the property to simplify or substitute.
- Simplify Expressions: An expression like sin(sin⁻¹(x) + cos⁻¹(x)) can be immediately simplified to sin(π/2), which equals 1.
- Convert Functions: It allows you to express one inverse function in terms of another, for example, writing sin⁻¹(x) as π/2 - cos⁻¹(x). This is crucial when an equation has a mix of sin⁻¹ and cos⁻¹ terms and you need to consolidate them.
4. Why is it essential to use the principal value branch when solving inverse trigonometric function problems?
It is essential because trigonometric functions are periodic, meaning they repeat their values at regular intervals. For example, sin(x) = 1/2 for x = π/6, 5π/6, 13π/6, etc. Consequently, their inverses are multi-valued. To ensure that an inverse trigonometric function provides a single, consistent output for a given input, its range is restricted to a specific interval called the principal value branch. This restriction makes the inverse a true function, which is fundamental for consistent calculations and proofs in trigonometry and calculus.
5. What is a common mistake when applying formulas for negative arguments, such as for cos⁻¹(-x) and tan⁻¹(-x)?
A very common mistake is treating all inverse functions the same for negative arguments. The correct properties are different based on the function's range:
- For sin⁻¹, tan⁻¹, and cosec⁻¹, the negative sign comes out directly: tan⁻¹(-x) = -tan⁻¹(x).
- For cos⁻¹, sec⁻¹, and cot⁻¹, the property is cos⁻¹(-x) = π - cos⁻¹(x).
The mistake is forgetting to include the 'π -' part for cos⁻¹, sec⁻¹, and cot⁻¹, and incorrectly writing cos⁻¹(-x) as -cos⁻¹(x). This error arises from not considering the principal value range for each function, which for cos⁻¹(x) is [0, π].
6. When solving an inverse trigonometry problem, how do I decide whether to use a substitution (e.g., x = tanθ) or apply a direct formula?
The choice of method depends on the structure of the problem. Here is a general guideline:
- Use Substitution: This is the best approach when the expression inside the inverse function resembles a standard trigonometric identity. For example, in an expression like tan⁻¹((2x)/(1-x²)), substituting x = tanθ transforms the inner part to tan(2θ), simplifying the problem significantly.
- Use Direct Formulas: Apply direct formulas, such as tan⁻¹(a) + tan⁻¹(b), when the problem involves the sum or difference of two or more simple inverse trigonometric terms. This method directly combines the terms into a single inverse function.
