Inverse Trigonometric Functions — Flashcards

Answer

Let $\theta = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$ $\Rightarrow \sin \theta = \frac{1}{\sqrt{2}}$ We know $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ Since $\frac{\pi}{4} \in \left[-\frac{\pi}{2…

Answer

Let $\theta = \cos^{-1}\left(-\frac{1}{2}\right)$ $\Rightarrow \cos \theta = -\frac{1}{2}$ We know $\cos \frac{2\pi}{3} = -\frac{1}{2}$ Since $\frac{2\pi}{3} \in [0, \pi]$, it lies in the principal…

Answer

Let $\theta = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right)$ $\Rightarrow \tan \theta = -\frac{1}{\sqrt{3}}$ We know $\tan \left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$ Since $-\frac{\pi}{6} \in \…

Answer

Using the property: $\cos(\cos^{-1}x) = x$ for $x \in [-1, 1]$ Here $\frac{1}{3} \in [-1, 1]$, so the property applies. $\therefore \cos\left(\cos^{-1}\frac{1}{3}\right) = \frac{1}{3}$ Answer: $\fr…

Answer

Using the property: $\cosec^{-1}(\cosec \theta) = \theta$ for $\theta \in \left[-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$ Here $\frac{\pi}{4} \in \left(0, \frac{\pi}{2}\right]$, so…

Answer

Let $\theta = \tan^{-1}x \Rightarrow \tan \theta = x$ Construct a right triangle: - Opposite = $x$, Adjacent = $1$, so Hypotenuse = $\sqrt{1 + x^2}$ $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adj…

Answer

Let $\theta = \sin^{-1}x \Rightarrow \sin \theta = x$ Using identity: $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - x^2}$ Since $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, $\cos \…

Answer

Let $\theta = \cosec^{-1}x \Rightarrow \cosec \theta = x$ Using identity: $\cot^2 \theta = \cosec^2 \theta - 1 = x^2 - 1$ $\Rightarrow \cot \theta = \sqrt{x^2 - 1}$ (taking positive root for princip…