Given that, $$\cos (\theta+\phi)=m \cos (\theta-\phi)$$

$$\Rightarrow \quad \frac{\cos (\theta+\phi)}{\cos (\theta-\phi)}=\frac{m}{1}$$

$$\begin{aligned} &\text { Using componendo and dividendo rule, }\\ &\begin{aligned} \frac{\cos (\theta-\phi)-\cos (\theta+\phi)}{\cos (\theta-\phi)+\cos (\theta+\phi)} & =\frac{1-m}{1+m} \\ \Rightarrow \frac{-2 \sin \left(\frac{\theta-\phi+\theta+\phi}{2}\right) \cdot \sin \left(\frac{\theta-\phi-\theta-\phi}{2}\right)}{2 \cos \left(\frac{\theta-\phi+\theta+\phi}{2}\right) \cdot \cos \left(\frac{\theta-\phi-\theta-\phi}{2}\right)} & =\frac{1-m}{1+m} \\ \Rightarrow \quad \frac{\sin \theta \cdot \sin \phi}{\cos \theta \cdot \cos \phi} & =\frac{1-m}{1+m} \quad \left[\begin{array}{lr} \because \quad \sin (-\theta) & =-\sin \theta \\ \text { and } \cos (-\theta) & =\cos \theta \end{array}\right]\\ \Rightarrow \quad \tan \theta \cdot \tan \phi & =\frac{1-m}{1+m} \\ \Rightarrow \quad \tan \theta & =\left(\frac{1-m}{1+m}\right) \cot \phi \end{aligned} \end{aligned}$$

Given expression,

$$\begin{gathered} 3\left[\sin ^4\left(\frac{3 \pi}{2}-\alpha\right)+\sin ^4(3 \pi+\alpha)\right]-2\left[\sin ^6\left(\frac{\pi}{2}+\alpha\right)+\sin ^6(5 \pi-\alpha)\right] \\ \quad=3\left[\cos ^4 \alpha+\sin ^4(\pi+\alpha)\right]-2\left[\cos ^6 \alpha+\sin ^6(\pi-\alpha)\right] \\ \quad=3\left[\cos ^4 \alpha+\sin ^4 \alpha\right]-2\left[\cos ^6 \alpha+\sin ^6 \alpha\right]=3-2=1 \end{gathered}$$

Given that, $a \cos 2 \theta+b \sin 2 \theta=c$

$$\Rightarrow a\left(\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right)+b\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)=c \quad\left[\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^2 \theta} \text { and } \cos 2 \theta=\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\right]$$

$$\begin{array}{ll} \Rightarrow & a\left(1-\tan ^2 \theta\right)+2 b \tan \theta=c\left(1+\tan ^2 \theta\right) \\ \Rightarrow & a-a-\tan ^2 \theta+2 b \tan \theta=c+c \tan ^2 \theta \\ \Rightarrow & (a+c) \tan ^2 \theta-2 b \tan \theta+c-a=0 \end{array}$$

Since, this equation has $\tan \alpha$ and $\tan \beta$ as its roots.

$$\because \quad \tan \alpha+\tan \beta=\frac{-(-2 b)}{a+c}=\frac{2 b}{a+c}$$

Given that, $$x=\sec \phi-\tan \phi\quad \text{.... (i)}$$

and $$y=\operatorname{cosec} \phi+\cot \phi\quad \text{.... (ii)}$$

Now, $\quad 1 \cdot x y=(\sec \phi-\tan \phi)(\operatorname{cosec} \phi+\cot \phi)$

$$\begin{aligned} & \Rightarrow \quad x y=\sec \phi \cdot \operatorname{cosec} \phi-\operatorname{cosec} \phi \cdot \tan \phi+\sec \phi \cdot \cot \phi-\tan \phi \cdot \cot \phi \\ & \Rightarrow \quad x y=\sec \phi \cdot \operatorname{cosec} \phi-\frac{1}{\cos \phi}+\frac{1}{\sin \phi}-1 \\ & \Rightarrow \quad 1+x y=\sec \phi \operatorname{cosec} \phi-\sec \phi+\operatorname{cosec} \phi\quad \text{.... (iii)} \end{aligned}$$

From Eqs. (i) and (ii), we get

$$\begin{array}{cc} & x-y=\sec \phi-\tan \phi-\operatorname{cosec} \phi-\cot \phi \\ \Rightarrow & x-y=\sec \phi-\operatorname{cosec} \phi-\frac{\sin \phi}{\cos \phi}-\frac{\cos \phi}{\sin \phi} \\ \Rightarrow & x-y=\sec \phi-\operatorname{cosec} \phi-\left(\frac{\sin ^2 \phi+\cos ^2 \phi}{\sin \phi \cdot \cos \phi}\right) \\ \Rightarrow & x-y=\sec \phi-\operatorname{cosec} \phi-\frac{1}{\sin \phi \cdot \cos \phi} \\ \Rightarrow & x-y=\sec \phi-\operatorname{cosec} \phi-\operatorname{cosec} \phi \cdot \sec \phi \\ \Rightarrow & x-y=-(\sec \phi \cdot \operatorname{cosec} \phi-\sec \phi+\operatorname{cosec} \phi) \\ \Rightarrow & x-y=-(x y+1) \quad \text{[from Eq. (iii)]}\\ \Rightarrow & xy+x-y+1=0\quad \text{Hence proved.} \end{array}$$

$\begin{aligned} &\begin{aligned} \begin{array}{ll} \text { Given that, }& \cos 3 \theta=\frac{8}{17} \Rightarrow \sin \theta=\sqrt{1-\frac{64}{289}} \\ \Rightarrow & \sin \theta=\sqrt{\frac{289-64}{289}} \Rightarrow \sin \theta= \pm \frac{15}{17} \\ \Rightarrow & \sin \theta=\frac{15}{17}\quad {[\text{since}, \theta ~\text{lies in first quadrant]}} \end{array} \end{aligned} \end{aligned}$

$$\begin{aligned} \text { Now, } \quad \cos (30 \Upsilon+\theta)+ & \cos (45 \Upsilon-\theta)+\cos (120 \Upsilon-\theta) \\ & =\cos (30 \Upsilon+\theta)+\cos (45 \Upsilon-\theta)+\cos (90 \Upsilon+30 \Upsilon-\theta) \\ & =\cos (30 \Upsilon+\theta)+\cos (45 \Upsilon-\theta)-\sin (30 \Upsilon-\theta) \\ & =\cos 30 \curlyvee \cos \theta-\sin 30 \curlyvee \sin \theta+\cos 45 \curlyvee \cos \theta+\sin 45 \Upsilon \sin \theta \\ & \quad-\sin 30 \curlyvee \cos \theta+\cos 30 \curlyvee \sin \theta \end{aligned}$$

$$\begin{aligned} & =\frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{2} \cos \theta \frac{\sqrt{3}}{2} \sin \theta \\ & =\left(\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}-\frac{1}{2}\right) \cos \theta+\left(\frac{1}{\sqrt{2}}-\frac{1}{2}+\frac{\sqrt{3}}{2}\right) \sin \theta \\ & =\left(\frac{\sqrt{6}+2-\sqrt{2}}{2 \sqrt{2}}\right) \cos \theta+\left(\frac{2-\sqrt{2}+\sqrt{6}}{2 \sqrt{2}}\right) \sin \theta \\ & =\left(\frac{\sqrt{6}+2-\sqrt{2}}{2 \sqrt{2}}\right) \frac{8}{17}+\left(\frac{2-\sqrt{2}+\sqrt{6}}{2 \sqrt{2}}\right) \frac{15}{17} \end{aligned}$$

$$\begin{aligned} & =\frac{1}{17(2 \sqrt{2})}(8 \sqrt{6}+16-8 \sqrt{2}+30-15 \sqrt{2}+15 \sqrt{6}) \\ & =\frac{1}{17(2 \sqrt{2})}(23 \sqrt{6}-23 \sqrt{2}+46) \\ & =\frac{23 \sqrt{6}}{17(2 \sqrt{2})}-\frac{23 \sqrt{2}}{17(2 \sqrt{2})}+\frac{46}{17(2 \sqrt{2})} \\ & =\frac{23 \sqrt{3}}{17(2)}-\frac{23}{17(2)}+\frac{23}{17 \sqrt{2}} \\ & =\frac{23}{17}\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right) \end{aligned}$$