Domain and range

Let \(y=\sec^{-1}x\). Then \(x=\sec y\). Since \(\sec y=\frac{1}{\cos y}\), we must have \(\cos y\ne 0\), and also \(|\sec y|\ge 1\). Therefore \(x\) must satisfy \(x\le -1\) or \(x\ge 1\).

Let \(y=\cosec^{-1}x\). Then \(x=\cosec y\). Since \(\cosec y=\frac{1}{\sin y}\), we must have \(\sin y\ne 0\), and also \(|\cosec y|\ge 1\). Therefore \(x\) must satisfy \(x\le -1\) or \(x\ge 1\).

Let \(y=\sin^{-1}x\). By definition, \(y\) is the angle whose sine equals \(x\), i.e., \(\sin y = x\). The sine function has range \([-1,1]\). So for \(\sin y\) to equal \(x\), the input \(x\) must satisfy \(-1 \le x \le 1\). Hence the domain of \(\sin^{-1}x\) is \([ -1,1 ]\).

Let \(y=\cos^{-1}x\). Then \(\cos y = x\). The cosine function has range \([-1,1]\). Therefore \(x\) can take values only in \([-1,1]\). Hence the domain of \(\cos^{-1}x\) is \([ -1,1 ]\).

The inverse tangent function tan⁻¹x is defined as the inverse of tangent after restricting tangent to its principal interval (−π/2, π/2). On that interval, tan y is continuous, strictly increasing, and takes every real value from −∞ to +∞. Therefore every real number x has exactly one principal angle y such that tan y = x. The input set, or domain, of tan⁻¹x is consequently all real numbers, ℝ. Option C is correct. The interval [−1, 1] is the domain of inverse sine and inverse cosine, not inverse tangent. The interval [0, π] is associated with the principal range of inverse cosine, while option D describes neither the domain nor the range of tan⁻¹x.

Direct answer: option B, it is not defined over the real numbers. For a real angle, cosine always lies between \(-1\) and 1, inclusive. Therefore the real domain of \(\cos^{-1}x\) is \([-1,1]\). Since \(-2<-1\), \(\cos^{-1}(-2)\) has no real value. A is false because the input is outside the domain. C is false: although \(\cos\pi=-1\), not \(-2\), so the value cannot be \(\pi\). D is false because \(\cos0=1\), not \(-2\). This is a domain check, so no calculation of a principal angle is possible. Memory cue: inverse sine and cosine accept only inputs from \(-1\) to 1.

Since \(\frac{7}{6}\notin\left[-1,1\right]\), \(\sin^{-1}\left(\frac{7}{6}\right)\) is not defined. Check the domain first.

The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\). Since \(\frac{11}{10}>1\), it is not defined.

For \(\sec^{-1}x\), we need \(\left|x\right|\ge1\). Here \(\left|\frac{3}{2}\right|\ge1\), so it is defined.

The function \(cosec^{-1}x\) is defined when \(\left|x\right|\ge1\). Here \(\left|-\frac{5}{4}\right|\ge1\), so it is defined.

The domain of arccos, \(\cos^{-1}\), is \([-1,1]\). Thus \(\cos^{-1}(-1)\) is defined and equals \(\pi\), so the expression equals \(\pi/\pi=1\). \(\cos^{-1}(3)\) and \(\sin^{-1}(5/4)\) are undefined because 3 and 5/4 lie outside the domain \([-1,1]\). Arcsec has domain \(|x|\ge 1\), so \(\sec^{-1}(1/2)\) is also undefined. Exam tip: always check the allowed domain of the inverse trig function before simplifying; values outside the domain make the expression invalid regardless of algebraic form.