---

Solution

---

Solution

---

Solution

---

Solution

---

Solution

Use identity: $\sin^{-1}x + \sin^{-1}y = \sin^{-1}\left(x\sqrt{1-y^2} + y\sqrt{1-x^2}\right)$ $= \sin^{-1}\left(\frac{8}{17}\cdot\frac{4}{5} + \frac{3}{5}\cdot\frac{15}{17}\right)$ $= \sin^{-1}\left(\frac{32}{85} + \frac{45}{85}\right)$ $= \sin^{-1}\left(\frac{77}{85}\right)$ Also, $\sin^{-1}x = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ $= \tan^{-1}\left(\frac{77/85}{\sqrt{1-(77/85)^2}}\right) = \tan^{-1}\left(\frac{77}{36}\right)$ $\boxed{\text{Both (a) and (b)}}$

---

Solution

We have, $\cos^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Using the identity $\cos^{-1} \sqrt{1 - p} = \sin^{-1} \sqrt{p}$:$\cos^{-1} \sqrt{p} + \sin^{-1} \sqrt{p} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$Since $\cos^{-1} \theta + \sin^{-1} \theta = \frac{\pi}{2}$:$\frac{\pi}{2} + \cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4}$$\cos^{-1} \sqrt{1 - q} = \frac{3\pi}{4} - \frac{\pi}{2}$$\cos^{-1} \sqrt{1 - q} = \frac{\pi}{4}$$\sqrt{1 - q} = \cos\left(\frac{\pi}{4}\right)$$\sqrt{1 - q} = \frac{1}{\sqrt{2}}$Squaring both sides:$1 - q = \frac{1}{2}$$q = 1 - \frac{1}{2}$$q = \frac{1}{2}$

---

Solution

Given, $\sin^{-1} x - \cos^{-1} x = \frac{\pi}{6}$ — (i)We know the identity: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ — (ii)Adding equations (i) and (ii):$(\sin^{-1} x - \cos^{-1} x) + (\sin^{-1} x + \cos^{-1} x) = \frac{\pi}{6} + \frac{\pi}{2}$$2\sin^{-1} x = \frac{\pi + 3\pi}{6}$$2\sin^{-1} x = \frac{4\pi}{6}$$2\sin^{-1} x = \frac{2\pi}{3}$$\sin^{-1} x = \frac{\pi}{3}$$x = \sin\left(\frac{\pi}{3}\right)$$x = \frac{\sqrt{3}}{2}$

---

Solution

$f'(x)=2\cos x(-\sin x)=-\sin 2x$ On $(0,\frac{\pi}{2})$, $\sin 2x > 0 \Rightarrow f'(x)<0$ So decreasing on $(0,\frac{\pi}{2})$

---

Solution

Check values in domain $[-1,1]$: $x=0,1,-1$ satisfy equation