Introduction to inverse trigonometric functions

Direct answer: option B, \(\frac{\pi}{2}\). The notation \(\sin^{-1}1\) means the principal angle whose sine is 1; it does not mean \(1/\sin1\). On the unit circle and from standard values, \(\sin(\pi/2)=1\). The principal range of inverse sine is \([-\pi/2,\pi/2]\), and \(\pi/2\) lies in it, so \(\sin^{-1}(1)=\pi/2\). Option A, 0, is wrong because \(\sin0=0\). Option C, \(\pi\), is wrong because \(\sin\pi=0\), and it is outside the principal range. Option D, \(-\pi/2\), has sine -1. The common mistake is confusing inverse function with reciprocal. Memory cue: inverse sine asks, ‘which principal angle has this sine?’

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 of inverse cosine, if \(y=\cos^{-1}x\), then \(y\) is the angle whose cosine is \(x\). Hence the correct relation is \(\cos y=x\).

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 \(\tan^{-1}x\).

\(\sec^{-1}(1/2)=\theta\) means \(\sec\theta=\frac{1}{2}\). So \(\cos\theta=\frac{1}{2}\). The principal value for \(\sec^{-1}\) is taken in \([0,\pi]\). In this interval, \(\cos\theta=\frac{1}{2}\) occurs at \(\theta=\frac{\pi}{3}\). Hence \(\sec^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\).

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 \(\left(0,\pi\right)\).

The domain of \(\sec^{-1}x\) is \(\left(-\infty,-1\right]\cup\left[1,\infty\right)\). The number (0) is not in this domain.

As (x\to\infty), (\tan^{-1}x\to\frac{\pi}{2}). The value (\frac{\pi}{2}) is a limit, not an attained value.

The principal range of tan⁻¹x is (−π/2, π/2). As x becomes increasingly negative, the angle in this range whose tangent equals x approaches the lower endpoint −π/2. In limit notation, lim x→−∞ tan⁻¹x = −π/2. Thus option A is correct. The value is approached from above; it is a horizontal asymptotic value rather than a value attained at a finite input. Option B is the corresponding limit when x→+∞, so its sign is wrong here. Option C would describe behavior near x = 0, not at negative infinity, and option D lies outside the principal range of arctangent. The sign and the direction of infinity are both essential in identifying the correct limit.

The value of (\sin x) lies only in ([-1,1]) so the domain of (\sin^{-1}x) is ([-1,1]). In exams check the domain first.

Let \(y=\cos^{-1}x\). Then \(\cos y=x\). Since cosine has range \([-1,1]\), the input \(x\) must lie in \([-1,1]\). Therefore the domain of \(\cos^{-1}x\) is \([ -1,1 ]\).