Introduction to inverse trigonometric functions

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.

The direct answer is B: \(\sin^{-1}x\) is the inverse function of sine, with the usual restricted sine domain used to make an inverse possible. Inverse means reversing an operation. If \(y=\sin^{-1}x\), then \(\sin y=x\), and the principal value of y is chosen in \([-\frac{\pi}{2},\frac{\pi}{2}]\). This is different from taking a reciprocal. Option A, \(\frac{1}{\sin x}\), is the reciprocal of \(\sin x\), called \(\cosec x\), not inverse sine. Option B is correct because it describes the inverse function: it takes a sine value as input and returns the corresponding principal angle. Option C is wrong because \(\cosec x=\frac{1}{\sin x}\), whereas inverse sine returns an angle; these are different kinds of quantities and are not always equal. Option D is wrong because inverse sine is defined for every real x in \([-1,1]\), not only for x=0. For example, \(\sin^{-1}(1)=\frac{\pi}{2}\). Memory cue: superscript -1 on a function may mean inverse, while reciprocal is written as \(1/\sin x\); read the notation carefully.

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}}).

The expression \(\tan^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\) is real only when \(1-x^2\ge 0\), so \(\sqrt{1-x^2}\) is defined. That requires \(x\in\left[-1,1\right]\). On this domain, the inverse-sine and the transformed inverse-tangent represent the same angle, so the identity holds.