Question

sin^-1 (-x)=??

sec^-1 (-x)=??

tan^-1(-x)=??

Cot^-1 (-x)=??

Cosec^-1 (-x)=??

explain in each case .

Answer 1

Answer on Question #50711 – Math – Differential Calculus | Equations

1.

$$\sin^{-1}(-x) = -\sin^{-1}(x).$$

$\sin^{-1}(x)$ is the inverse of $\sin(x)$, it is odd function.

2.

$$\sec^{-1}(-x) = \pi - \sec^{-1}(x).$$

$\cosec^{-1}(x)$ is odd function ($\cosec^{-1}(x) = \sin^{-1}\left(\frac{1}{x}\right)$).

And

$$\sec^{-1}(-x) + \cosec^{-1}(-x) = \frac{\pi}{2} \rightarrow \sec^{-1}(-x) = \frac{\pi}{2} - \cosec^{-1}(-x) = \frac{\pi}{2} + \cosec^{-1}(x).$$

But

$$\cosec^{-1}(x) = \frac{\pi}{2} - \sec^{-1}(x).$$

Thus

$$\sec^{-1}(-x) = \frac{\pi}{2} + \frac{\pi}{2} - \sec^{-1}(x) = \pi - \sec^{-1}(x).$$

3.

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

$\tan^{-1}(x)$ is odd function.

4.

$$\cot^{-1}(-x) = \pi - \cot^{-1}(x).$$ $$\cot^{-1}(-x) + \tan^{-1}(-x) = \frac{\pi}{2} \rightarrow \cot^{-1}(-x) = \frac{\pi}{2} - \tan^{-1}(-x) = \frac{\pi}{2} + \tan^{-1}(x).$$

But

$$\tan^{-1}(x) = \frac{\pi}{2} - \cot^{-1}(x).$$

Thus

$$\cot^{-1}(-x) = \frac{\pi}{2} + \frac{\pi}{2} - \cot^{-1}(x) = \pi - \cot^{-1}(x).$$

5.

$$\cosec^{-1}(-x) = -\cosec^{-1}(x).$$

$\cosec^{-1}(x)$ is odd function ($\cosec^{-1}(x) = \sin^{-1}\left(\frac{1}{x}\right)$).

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