Introduction to inverse trigonometric functions

The function \(\tan^{-1}x\) is defined for every real (x). In the other options the domain condition fails.

The domain of \(\tan^{-1}x\) is \(\mathbb{R}\) and its range is \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\). The endpoints of the open interval are not included.

By making (\cos x) one-one on \(\left[0,\pi\right]\), \(\cos^{-1}x\) becomes decreasing on \(\left[-1,1\right]\). The graph direction also confirms this.

The function \(\sin^{-1}x\) is increasing on its whole domain \(\left[-1,1\right]\). This matches the chosen branch of (\sin x).

Restricting (\sin x) to (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]) makes it one-one and gives values (\left[-1,1\right]). This defines (\sin^{-1}x).

The function (\cos x) is made one-one on (\left[0,\pi\right]). Therefore the range of (\cos^{-1}x) is (\left[0,\pi\right]).

Here (\sin^{-1}1=\frac{\pi}{2}) and (\cos^{-1}1=0), so the sum is (\frac{\pi}{2}). Remember endpoint values.

Since (\cos^{-1}0=\frac{\pi}{2}) and (\sin^{-1}0=0), the difference is (\frac{\pi}{2}). Inverse trigonometric values at zero are very useful.

Here (\tan^{-1}0=0) and (\cot^{-1}0=\frac{\pi}{2}), so the sum is (\frac{\pi}{2}). Do not write (\cot^{-1}0) as (0).

From (\sin^{-1}x=\frac{\pi}{4}), (x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}). Convert the inverse form into the normal trigonometric form.

From (\cos^{-1}x=\frac{2\pi}{3}), (x=\cos\frac{2\pi}{3}=-\frac{1}{2}). Decide the sign from the quadrant of the angle.

From \(\tan^{-1}x=-\frac{\pi}{6}\), \(x=\tan\left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}\). The range of \(\tan^{-1}\) also includes negative angles.

From (\sec y=-\sqrt{2}), (\cos y=-\frac{1}{\sqrt{2}}), so the principal value is (y=\frac{3\pi}{4}). For (\sec^{-1}), decide the quadrant from the sign.

The domain of \(\cos^{-1}x\) is \(\left[-1,1\right]\) and its range is \(\left[0,\pi\right]\). In domain-range questions, identify intervals first.

Since (\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}), if both are equal then each is (\frac{\pi}{4}). Hence (x=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}).

For (x>0), (\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}). Always check the range convention of (\cot^{-1}x).

For every (x\in[-1,1]), (\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}). Therefore the equation with (\pi) is impossible.

(\sin^{-1}x) is inverse sine, not reciprocal. Keep the difference between (\sin^{-1}x) and ((\sin x)^{-1}).

For every (a\in[-1,1]), (\sin^{-1}a+\cos^{-1}a=\frac{\pi}{2}). They are equal only at (a=\frac{1}{\sqrt{2}}).

Let \(\theta=\sin^{-1}x\) so that \(\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\) and \(\sin\theta=x\). In this principal range \(\cos\theta=\sqrt{1-x^2}\) (nonnegative), hence
\(\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{x}{\sqrt{1-x^2}}\).
Thus \(\tan^{-1}\big(\dfrac{x}{\sqrt{1-x^2}}\big)=\theta=\sin^{-1}x\) provided \(\sqrt{1-x^2}\neq0\) and \(\sin^{-1}x\) is defined — i.e. \(-1<x<1\). At \(x=\pm1\) the denominator is zero \(the expression is not defined even though the limit gives ±\(\tfrac{\pi}{2}\)\), and for \(|x|>1\) the left side \(\sin^{-1}x\) is not defined. Exam tip: always check domains and principal-value ranges when equating inverse trig functions; note the sign of the square root in the principal interval.