The inverse cosecant function is defined only for: x ≤ −1 or x ≥ 1. Thus values strictly between −1 and 1 are excluded from the domain. Among the options, 1/2 lies inside the forbidden interval (−1,1). Therefore cosec⁻¹(1/2) is undefined. Hence option B is correct.

Inverse trigonometric functions require restricted domains so the original function becomes bijective. Hence the assertion is true. However, the principal range of cosec⁻¹x is: [-π/2, π/2] − {0}, not [-1,1]. Therefore the reason is false. Hence option B is correct.

We seek θ such that: cosecθ=−√2. Since: sinθ=−1/√2, the principal value is: θ=−π/4. This lies in: [-π/2, π/2]−{0}, the principal branch of cosec⁻¹x. Hence option A is correct.

Natural domain of cosec x includes all real numbers except integral multiples of π. The domain of cosec⁻¹x is narrower: R−(−1,1). Its principal range: [-π/2,π/2]−{0} is even more restricted. Therefore correct ordering is: 1,2,3. Hence option C is correct.

The domain of sec⁻¹x is: x ≤ −1 or x ≥ 1. Thus all values between −1 and 1, including 0, are excluded. Therefore statement D is incorrect. Statements A, B, and C are mathematically correct. Hence option D is correct.

For sec⁻¹x: range is [0,π]−{π/2}. As x→∞, the corresponding angle approaches π/2 because: sec(π/2) becomes unbounded. However π/2 itself is excluded. Therefore the curve asymptotically approaches π/2. Hence option D is correct.

The principal range of sec⁻¹x is: [0,π]−{π/2}. Now: sec0=1 ⇒ sec⁻¹(1)=0, secπ=−1 ⇒ sec⁻¹(−1)=π. Both outputs belong to the principal range. Therefore both statements are correct. Hence option C is correct.

For principal branches: cosec⁻¹ excludes 0 from its range, sec⁻¹ excludes π/2, tan⁻¹ has continuous range: (−π/2,π/2). Hence matching becomes: 1-b, 2-c, 3-a. Therefore option D is correct.

The inverse tangent function has domain: R. A normally distributed variable can take any real value, and tan⁻¹x is defined for every real number. Therefore every possible draw produces a valid real output. Hence probability equals 1. Option D is correct.

The principal range of tan⁻¹x is: (−π/2,π/2). As x increases indefinitely, tan⁻¹x approaches π/2 asymptotically. The graph gets closer and closer to π/2 without touching it. Therefore the limiting value is π/2. Hence option C is correct.

The inverse tangent function has principal range: (-π/2, π/2). Since: tan(π/4)=1, the principal value becomes: tan⁻¹(1)=π/4. This equality is always true, so the probability is 1. Hence option B is correct.

The tangent function has period π. Hence inverse tangent values from adjacent branches differ by π. At x=1: principal branch gives π/4, adjacent branch gives 5π/4. Difference: 5π/4−π/4=π. Therefore option A is correct.

For cot⁻¹x, the principal range is: (0,π). We require an angle θ satisfying: cotθ=0. Since: cot(π/2)=0, the principal value is: π/2. Thus option C is correct.

Evaluate each principal value: tan⁻¹(1)=π/4, cos⁻¹(−1/2)=2π/3, sin⁻¹(−1/2)=−π/6. Adding: π/4 + 2π/3 − π/6 = π/4 + π/2 = 3π/4. Hence option A is correct.

We seek θ in the principal range: (0,π), such that: cotθ=−1. Since: cot(3π/4)=−1, the principal value is: 3π/4. Thus option B is correct.

Cotangent becomes bijective on any interval of length π avoiding discontinuities. Therefore intervals like: (0,π), (π,2π), (−π,0) all produce valid inverse branches. The reason correctly explains why multiple inverse branches exist. Hence option C is correct.

According to the NCERT graphs: List I — List II 1. Horizontal asymptotes at \(y=\pm \frac{\pi }{2}\) — b. \(y={tan⁡}^{-1}x\) 2. Horizontal asymptotes at \(y=0\) and \(y=\pi\) — a. \(y={cot⁡}^{-1}x\) 3. Strictly increasing function — d. Graph increases from left to right 4. Strictly decreasing function — c. Graph decreases from left to right The graph of \(y={tan⁡}^{-1}x\) increases continuously from left to right and approaches \(y=-\frac{\pi }{2}\)as \(x\rightarrow -\infty\) and \(y=\frac{\pi }{2}\)as \(x\rightarrow \infty\). The graph of \(y={cot⁡}^{-1}x\) decreases continuously from left to right and approaches \(y=\pi\) as \(x\rightarrow -\infty\) and \(y=0\) as \(x\rightarrow \infty\). Therefore, the correct matching is: 1 → b 2 → a 3 → d 4 → c Hence, Option A is correct.

Evaluate each quantity: sin⁻¹(0)=0, cot⁻¹(0)=π/2, cos⁻¹(−1)=π. Thus ordering from smallest to largest becomes: 0 < π/2 < π. Hence arrangement: 1, 2, 3. Therefore option D is correct.

The passage explicitly states: sin⁻¹ has domain: [-1,1] and range: [-π/2,π/2]. Therefore the correct domain of the principal inverse sine function is: [-1,1]. Hence option A is correct.

The passage clearly defines: "The value of an inverse trigonometric function which lies in the range of principal branch is called the principal value." Therefore the required term is: Principal value. Hence option A is correct.