NCERT Solutions Class 12 Maths Inverse Trigonometric Functions Ex 2.1

Core Mathematical Background: Why Do We Restrict Domains?

To construct inverse functions, a function must be a bijection. Since standard trigonometric functions like sine, cosine, and tangent repeat their values periodically, they are many-to-one. By restricting their domains to specific intervals, we ensure each output value corresponds to exactly one input value. For example, the domain of the sine function is restricted to [-π/2, π/2] to define its inverse function, sin⁻¹(x). The resulting values of these inverse functions are called principal values, and they must strictly lie within their respective Principal Value Branches. If your calculated angle falls outside this specified range, your solution is mathematically invalid for this exercise.

Principal Value Branches (PVB) of Inverse Trigonometric Functions

sin⁻¹(x)

Domain is [-1, 1] and the Principal Value Branch (Range) is [-π/2, π/2].

cos⁻¹(x)

Domain is [-1, 1] and the Principal Value Branch (Range) is [0, π].

tan⁻¹(x)

Domain is R (all real numbers) and the Principal Value Branch (Range) is (-π/2, π/2).

cosec⁻¹(x)

Domain is R - (-1, 1) and the Principal Value Branch (Range) is [-π/2, π/2] - {0}.

Step-by-Step Method to Find Principal Values

1. Step 1: Set the Expression Equal to y — Equate the given inverse trigonometric expression to a variable, say y. For example, let y = sin⁻¹(x).
2. Step 2: Express in Standard Trigonometric Form — Rewrite the equation as sin(y) = x, or the respective trigonometric form.
3. Step 3: Analyze the Sign of x — If x is positive, find the basic acute angle in the first quadrant. If x is negative, use mathematical identities to adjust the angle to the correct principal value branch range.
4. Step 4: Apply Negative Argument Rules — Use the formulas: sin⁻¹(-x) = -sin⁻¹(x), tan⁻¹(-x) = -tan⁻¹(x), and cosec⁻¹(-x) = -cosec⁻¹(x). For the other group, use: cos⁻¹(-x) = π - cos⁻¹(x), sec⁻¹(-x) = π - sec⁻¹(x), and cot⁻¹(-x) = π - cot⁻¹(x).

CBSE Exam Strategy & Common Traps

1. Never Confuse Inverse with Reciprocal: Remind yourself that sin⁻¹(x) is NOT equal to 1/sin(x). The notation sin⁻¹(x) denotes the angle whose sine is x, whereas (sin x)⁻¹ is cosec(x).
2. The Cosine Trap: When solving for cos⁻¹(-x), a common board-exam mistake is writing the answer as negative (like -π/3). Remember that the range of cos⁻¹(x) is [0, π]. Negative angles do not exist in this branch. Always subtract from π: cos⁻¹(-1/2) = π - π/3 = 2π/3.
3. Check the Brackets: Be mindful of open vs. closed intervals. For instance, the principal range of tan⁻¹(x) is the open interval (-π/2, π/2), which excludes -π/2 and π/2, whereas sin⁻¹(x) is the closed interval [-π/2, π/2] which includes them.

Practice Questions with Solutions

• Q: Find the principal value of sin⁻¹(-1/2). A: Step 1: Let y = sin⁻¹(-1/2). This can be rewritten as sin(y) = -1/2. Step 2: We know the principal value branch of sin⁻¹ is [-π/2, π/2]. Step 3: Since sin(π/6) = 1/2, and using the identity sin(-θ) = -sin(θ), we get sin(-π/6) = -1/2. Step 4: Since -π/6 lies within the range [-π/2, π/2], the value is valid. Final answer: The principal value of sin⁻¹(-1/2) is -π/6.
• Q: Find the principal value of cos⁻¹(-1/2). A: Step 1: Let y = cos⁻¹(-1/2). This can be rewritten as cos(y) = -1/2. Step 2: The principal value branch of cos⁻¹ is [0, π]. Step 3: We know that cos(π/3) = 1/2. To find the negative value within [0, π], we use the identity cos(π - θ) = -cos(θ). Step 4: cos(π - π/3) = cos(2π/3) = -1/2. Final answer: The principal value of cos⁻¹(-1/2) is 2π/3.
• Q: Find the principal value of tan⁻¹(-√3). A: Step 1: Let y = tan⁻¹(-√3). This can be rewritten as tan(y) = -√3. Step 2: The principal value branch of tan⁻¹ is (-π/2, π/2). Step 3: Since tan(π/3) = √3, and using the property tan(-θ) = -tan(θ), we get tan(-π/3) = -√3. Step 4: Since -π/3 lies in open interval (-π/2, π/2), it is the correct angle. Final answer: The principal value of tan⁻¹(-√3) is -π/3.
• Q: Find the value of the expression: tan⁻¹(1) + cos⁻¹(-1/2) + sin⁻¹(-1/2). A: Step 1: Find the individual principal values. Step 2: For tan⁻¹(1), let y = tan⁻¹(1) => tan(y) = 1 => y = π/4 (since π/4 is in (-π/2, π/2)). Step 3: For cos⁻¹(-1/2), we calculated this as 2π/3 in previous examples. Step 4: For sin⁻¹(-1/2), we calculated this as -π/6 in previous examples. Step 5: Substitute these values into the expression: π/4 + 2π/3 + (-π/6) = (3π + 8π - 2π) / 12 = 9π/12 = 3π/4. Final answer: The value of the expression is 3π/4.

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