1. **Problem Statement:** Evaluate the given inverse trigonometric expressions and trigonometric values involving inverse functions. 2. **Recall important formulas and rules:** - $\sin^{-1}(x)$ gives an angle $\theta$ such that $\sin \theta = x$ and $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. - $\cos^{-1}(x)$ gives an angle $\theta$ such that $\cos \theta = x$ and $\theta \in [0, \pi]$. - $\sec^{-1}(x)$ gives an angle $\theta$ such that $\sec \theta = x$ and $\theta \in [0, \pi]$, $\theta \neq \frac{\pi}{2}$. - $\cot^{-1}(x)$ gives an angle $\theta$ such that $\cot \theta = x$ and $\theta \in (0, \pi)$. - For expressions like $\sin[a - \sin^{-1}(b)]$, use angle subtraction formulas. - For $\sin^{-1}[\sin(\alpha)]$, the value is $\alpha$ if $\alpha$ is in the principal range of $\sin^{-1}$, otherwise adjust by $\pi$. --- **Step 1: Evaluate each part of Exercise 8(a):** (i) $\sin^{-1}(\frac{\sqrt{3}}{2})$: - $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ - Since $\frac{\pi}{3} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, answer is $\frac{\pi}{3}$. (ii) $\cos^{-1}(\frac{1}{\sqrt{2}})$: - $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ - $\frac{\pi}{4} \in [0, \pi]$, so answer is $\frac{\pi}{4}$. (iii) $\sec^{-1}(-\sqrt{2})$: - $\sec \theta = -\sqrt{2} \implies \cos \theta = -\frac{1}{\sqrt{2}}$ - $\cos \frac{3\pi}{4} = -\frac{1}{\sqrt{2}}$ - $\sec^{-1}$ range is $[0, \pi]$ excluding $\frac{\pi}{2}$, so answer is $\frac{3\pi}{4}$. (iv) $\cot^{-1}(-\sqrt{3})$: - $\cot \theta = -\sqrt{3}$ - $\cot \frac{5\pi}{6} = -\sqrt{3}$ (since $\cot \frac{\pi}{6} = \sqrt{3}$ and cotangent is negative in second quadrant) - $\cot^{-1}$ range is $(0, \pi)$, so answer is $\frac{5\pi}{6}$. (v) $\sin \left[ \frac{\pi}{3} - \sin^{-1}(-\frac{1}{2}) \right]$: - $\sin^{-1}(-\frac{1}{2}) = -\frac{\pi}{6}$ (since $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$) - Expression becomes $\sin \left( \frac{\pi}{3} - (-\frac{\pi}{6}) \right) = \sin \left( \frac{\pi}{3} + \frac{\pi}{6} \right) = \sin \frac{\pi}{2} = 1$. (vi) $\sin^{-1}[\sin(\frac{5\pi}{6})]$: - $\sin \frac{5\pi}{6} = \frac{1}{2}$ - $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ (principal value in $[-\frac{\pi}{2}, \frac{\pi}{2}]$) - So answer is $\frac{\pi}{6}$. --- **Step 2: Evaluate the second set:** (i) $\sin(\cos^{-1} \frac{3}{5})$: - Let $\theta = \cos^{-1} \frac{3}{5}$, so $\cos \theta = \frac{3}{5}$. - Using $\sin^2 \theta + \cos^2 \theta = 1$, we get $$\sin \theta = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}.$$ - Since $\theta \in [0, \pi]$, $\sin \theta \geq 0$, so answer is $\frac{4}{5}$. (ii) $\tan(\csc^{-1} \frac{65}{63})$: - Let $\alpha = \csc^{-1} \frac{65}{63}$, so $\csc \alpha = \frac{65}{63}$. - Then $\sin \alpha = \frac{63}{65}$. - Using $\sin^2 \alpha + \cos^2 \alpha = 1$, $$\cos \alpha = \sqrt{1 - \left(\frac{63}{65}\right)^2} = \sqrt{1 - \frac{3969}{4225}} = \sqrt{\frac{256}{4225}} = \frac{16}{65}.$$ - Then $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{63/65}{16/65} = \frac{63}{16}$. (iii) $\cos^{-1}[\cos(\frac{5\pi}{4})]$: - $\cos \frac{5\pi}{4} = -\frac{1}{\sqrt{2}}$. - $\cos^{-1}(-\frac{1}{\sqrt{2}})$ is the angle in $[0, \pi]$ with cosine $-\frac{1}{\sqrt{2}}$, which is $\frac{3\pi}{4}$. --- **Final answers:** 1.(i) $\frac{\pi}{3}$ 1.(ii) $\frac{\pi}{4}$ 1.(iii) $\frac{3\pi}{4}$ 1.(iv) $\frac{5\pi}{6}$ 1.(v) $1$ 1.(vi) $\frac{\pi}{6}$ 2.(i) $\frac{4}{5}$ 2.(ii) $\frac{63}{16}$ 2.(iii) $\frac{3\pi}{4}$