Introduction to inverse trigonometric functions

Since (\sin\frac{\pi}{2}=1) and (\frac{\pi}{2}) is in the principal range, the answer is (\frac{\pi}{2}).

Since \(\cot\frac{\pi}{2}=0\) and \(\frac{\pi}{2}\) lies in \(\left(0,\pi\right)\), the value is \(\frac{\pi}{2}\).

\(\tan^{-1}x\) is defined for every real number; for example, \(x=2\) is allowed. In contrast, \(\sin^{-1}x\) requires \(-1\le x\le1\). Exam tip: check the domain first.

The expression (\sin^{-1}x) gives the angle whose (\sin) value is (x). So (y=\sin^{-1}x) means (\sin y=x).

By definition, \(y=\cos^{-1}x\) means y is the angle whose cosine equals \(x\). Hence \(\cos y = x\) is correct. The other options are not generally true: \(\sin y=x\) holds only for special values of \(x\), and \(\tan y\) or \(\cot y\) are ratios involving \(\sin y\) and \(\cos y\), so they are not simply equal to \(x\). Exam tip: remember the principal value/range of \(\cos^{-1}x\) is \([0,\pi]\).

By definition, \(y=\tan^{-1}x\) means y is the angle whose tangent equals x; hence \(\tan y = x\). The other choices (\(\sin y = x\), \(\cos y = x\), \(\sec y = x\)) are not generally true for the angle given by \(\tan^{-1}x\) except in special cases, so they are not correct in general. Exam tip: Memorize the definitions and principal value ranges of inverse trig functions (e.g. \(-\pi/2<y<\pi/2\) for \(\tan^{-1}x\)).

The domain of \(\sin^{-1}x\) is \(\left[-1,1\right]\) and \(2\notin\left[-1,1\right]\). Hence it is not defined.

The inverse tangent function \(\tan^{-1}x\) is defined for every real number \(x\in\mathbb{R}\) and its principal value lies in the interval \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\). Hence \(\tan^{-1}5\) is defined and equals a finite angle inside that open interval. The choice \(\tfrac{\pi}{2}\) is incorrect because \(\tan\theta\) has a vertical asymptote at \(\theta=\tfrac{\pi}{2}\) and arctan never attains that endpoint; \(\pi\) is also impossible since arctan’s principal value cannot be \(\pi\). Exam tip: memorize domain = all real numbers and range = \((-\tfrac{\pi}{2},\tfrac{\pi}{2})\) for \(\tan^{-1}x\).

The inverse secant \(\sec^{-1}x\) is defined only for \(|x|\ge 1\), because \(\sec\theta=1/\cos\theta\) and therefore \(|\sec\theta|\ge1\). Here \(\left|\tfrac{1}{2}\right|<1\), so \(\sec^{-1}\left(\tfrac{1}{2}\right)\) is not defined. Options C and D are incorrect: \(\sec\theta=\tfrac{1}{2}\) would require \(\cos\theta=2\), which is impossible. \(For reference, \(\sec(\pi/3)=2\), so \(\sec^{-1}(2)=\pi/3\), but that does not imply \(\sec^{-1}(1/2)\) exists.\) Exam tip: always check the domain of the inverse trigonometric function first — for arcsec the requirement is \(|x|\ge1\).

The function (\sin^{-1}x) is the inverse of (\sin x) on a suitable branch. Hence it is an inverse trigonometric function.

The notation (\sin^{-1}x) means the inverse function of (\sin x), not the reciprocal (\frac{1}{\sin x}). This is an important mistake to avoid.

The identity \(\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\) is true for all \(x\in\left[-1,1\right]\). Remember this basic identity for exams.

The identity (\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}) is used for all real (x). This is a very useful introductory identity.

The statement \(\theta=sin^{-1}\left(\frac{3}{5}\right)\) means \(sin\theta=\frac{3}{5}\). Apply the definition directly.

The expression \(\cos^{-1}\left(\frac{4}{5}\right)\) is the angle whose cosine is \(\frac{4}{5}\). So \(\cos\theta=\frac{4}{5}\).

The statement \(\theta=\tan^{-1}\left(\frac{5}{12}\right)\) means \(\tan\theta=\frac{5}{12}\). Remember the definition of inverse function.

Given \(\tan\theta=\dfrac{8}{15}\), take opposite = 8 and adjacent = 15 in a right triangle. The hypotenuse is \(\sqrt{8^2+15^2}=\sqrt{289}=17\). Thus \(\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}}=\dfrac{15}{17}\). Closest distractor \(\dfrac{8}{17}\) is actually \(\sin\theta\); the options \(\dfrac{17}{15}\) and \(\dfrac{17}{8}\) are reciprocals greater than 1 and cannot be cosine values for a real acute angle. Exam tip: Recognize the 8–15–17 Pythagorean triple to compute hypotenuse quickly and avoid arithmetic errors.

The principal inverse cosine function has domain [−1, 1] and range [0, π]. As x increases from −1 to 1, the angle whose cosine is x decreases from π to 0. For example, cos⁻¹(−1) = π, cos⁻¹(0) = π/2, and cos⁻¹(1) = 0. These values show the decreasing behavior, and the monotonicity is strict throughout the domain. Therefore option B is correct. It is not increasing because larger cosine inputs correspond to smaller principal angles. It is not constant because the output changes continuously from π to 0. The function is defined for every x in [−1, 1], so option D is also incorrect. This conclusion follows from the restricted principal branch of cosine.

The function \(\tan^{-1}x\) is defined and increasing for all real numbers. Its range is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\).

The function \(\cot^{-1}x\) is decreasing on \(\mathbb{R}\). Its principal value range is \(\left(0,\pi\right)\).