The given equations are

$$\begin{aligned} & \tan \theta=-1 \quad \text{.... (i)}\\ \text { and } & \cos \theta=\frac{1}{\sqrt{2}} \quad \text{.... (ii)}\\ \text{From Eq. (i),}\quad & \tan \theta=-\tan \frac{\pi}{4} \\ \Rightarrow \quad & \tan \theta=\tan \left(2 \pi-\frac{\pi}{4}\right) \Rightarrow \tan \theta=\tan \frac{7 \pi}{4} \\ \therefore \quad & \theta=\frac{7 \pi}{4} \end{aligned}$$

From Eq. (ii),

$$\begin{aligned} & \cos \theta=\frac{1}{\sqrt{2}} \Rightarrow \cos \theta=\cos \frac{\pi}{4} \\ & \Rightarrow \quad \cos \theta=\cos \left(2 \pi-\frac{\pi}{4}\right) \Rightarrow \cos \theta=\cos \frac{7 \pi}{4} \\ & \therefore \quad \theta=\frac{7 \pi}{4} \end{aligned}$$

Hence, the most general value of $\theta$ i.e., $\theta=2 n \pi+\frac{7 \pi}{4}$.

$$\begin{aligned} & \text { Given that, } \quad \cot \theta+\tan \theta=2 \operatorname{cosec} \theta \\ & \Rightarrow \quad \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=\frac{2}{\sin \theta} \\ & \Rightarrow \quad \frac{\cos ^2+\sin ^2 \theta}{\sin \theta \cdot \cos \theta}=\frac{2}{\sin \theta} \\ & \Rightarrow \quad \frac{1}{\cos \theta}=2 \quad [\because \sin^2\theta+\cos^2\theta=1]\\ & \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\cos \frac{\pi}{3} \\ & \therefore \quad \theta=2 n \pi \pm \frac{\pi}{3} \end{aligned}$$

$$\begin{aligned} \text{Given that,}\quad & 2 \sin ^2 \theta =3 \cos \theta \\ \Rightarrow \quad & 2-2 \cos ^2 \theta =3 \cos \theta \\ \Rightarrow \quad & 2 \cos ^2 \theta+3 \cos \theta- =0 \\ \Rightarrow \quad & 2 \cos ^2 \theta+4 \cos \theta-\cos \theta-2 =0 \\ \Rightarrow \quad & 2\cos \theta(\cos \theta+2)-1(\cos \theta+2) =0 \\ \Rightarrow \quad & (\cos \theta+2)(2 \cos \theta-1) =0 \\ \Rightarrow \quad & \cos \theta =-2 \text { not possible } \quad [\because -1\le \cos\theta \le 1]\\ & 2 \cos \theta =1 \\ \Rightarrow \quad & \cos \theta =\frac{1}{2} \\ \Rightarrow \quad & \cos \theta =\cos \frac{\pi}{3} \\ \therefore \quad & \theta =\frac{\pi}{3} \\ \text{Also,}\quad & \cos \theta =\cos \left(2 \pi-\frac{\pi}{3}\right) \\ \Rightarrow \quad &\cos \theta =\cos \frac{5 \pi}{6} \\ \therefore \quad &\theta =\frac{5 \pi}{6} \end{aligned}$$

So, the values of $\theta$ are $\frac{\pi}{3}$ and $\frac{5\pi}{6}$.

Given that,

$$\begin{aligned} & \sec x \cos 5 x+1=0 \\ & \frac{\cos 5 x}{\cos x}+1=0 \Rightarrow \cos 5 x+\cos x=0 \end{aligned}$$

$\Rightarrow \quad 2 \cos \left(\frac{5 x+x}{2}\right) \cdot \cos \left(\frac{5 x-x}{2}\right)=0 \quad\left[\because \cos x+\cos y=2 \cos \frac{x+y}{2} \cdot \cos \frac{x-y}{2}\right]$

$$ \begin{array}{lrl} \Rightarrow & 2 \cos 3 x \cdot \cos 2 x & =0 \\ \Rightarrow & \cos 3 x & =0 \text { or } \cos 2 x=0 \\ \Rightarrow & \cos 3 x & =\cos \frac{\pi}{2} \text { or } \cos 2 x=\cos \frac{\pi}{2} \\ \therefore & 3 x & =\frac{\pi}{2} \Rightarrow 2 x=\frac{\pi}{2} \\ \text { and } & x & =\frac{\pi}{6} \Rightarrow x=\frac{\pi}{4} \end{array} $$ Hence, the solutions are $\frac{\pi}{2}, \frac{\pi}{4}$ and $\frac{\pi}{6}$.

Given that, $\sin(\theta+\alpha)=a\quad\text{... (i)}$

and $\sin(\theta+\beta)=b\quad \text{... (ii)}$

$$\begin{array}{ll} \therefore & \cos (\theta+\alpha)=\sqrt{1-a^2} \text { and } \cos (\theta+\beta)=\sqrt{1-b^2} \\ \therefore & \cos (\alpha-\beta)=\cos \{\theta+\alpha-(\theta+\beta)\} \end{array}$$

$$\begin{aligned} & =\cos (\theta+\beta) \cos (\theta+\alpha)+\sin (\theta+\alpha) \sin (\theta+\beta) \\ & =\sqrt{1-a^2} \sqrt{1-b^2}+a \cdot b=a b+\sqrt{\left(1-a^2\right)\left(1-b^2\right)} \\ & =a b+\sqrt{1-a^2-b^2+a^2 b^2} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text{and}\quad & \cos (\alpha-\beta)=a b+\sqrt{1-a^2-b^2+a^2 b^2} \\ & =\cos 2(\alpha-\beta)-4 a b \cos (\alpha-\beta) \\ & =2 \cos ^2(\alpha-\beta)-1-4 a b \cos (\alpha-\beta) \\ & =2 \cos (\alpha-\beta)(\cos \alpha-\beta-2 a b)-1 \\ & =2\left(a b+\sqrt{1-a^2-b^2+a^2 b^2}\right)\left(a b+\sqrt{1-a^2-b^2+a^2 b^2}-2 a b\right)-1 \\ & =2\left[\left(\sqrt{1-a^2-b^2+a^2 b^2+a b}\right)\left(\sqrt{1-a^2-b^2+a^2 b^2}-a b\right)\right]-1 \\ & =2\left[1-a^2-b^2+a^2 b^2-a^2 b^2\right]-1 \\ & =2-2 a^2-2 b^2-1 \\ & =1-2 a^2-2 b^2\quad \text{Hence proved.} \end{aligned}\\ \end{aligned}$$