Given that $\tan (A+B)=p\quad \text{... (i)}$

and $\tan (A-B)=q\quad \text{.... (ii)}$

$\begin{aligned} \therefore \quad \tan 2 A & =\tan (A+B+A-B) \\ & =\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \tan (A-B)} \quad\left[\because \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right] \\ & =\frac{p+q}{1-p q} \quad \text { [from Eqs. (i) and (ii)] }\end{aligned}$

$\begin{aligned} & \text { Given that, } \quad \cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta \\ & \Rightarrow \quad(\cos \alpha+\cos \beta)^2-(\sin \alpha+\sin \beta)^2=0 \\ & \Rightarrow \quad \cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta-\sin ^2 \alpha-\sin ^2 \beta-2 \sin \alpha \sin \beta=0 \\ & \Rightarrow \quad \cos ^2 \alpha-\sin ^2 \alpha+\cos ^2 \beta-\sin ^2 \beta=2(\sin \alpha \sin \beta-\cos \alpha \cos \beta) \\ & \Rightarrow \quad \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)\quad \text{Hence proved.}\end{aligned}$

Given that, $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$

Using componendo and dividendo,

$$\begin{aligned} & \Rightarrow \quad=\frac{\sin (x+y)+[\sin (x-y)]}{\sin (x+y)-\sin (x-y)}=\frac{a+b+a-b}{a+b-a+b} \\ & \Rightarrow \quad=\frac{2 \sin \left(\frac{x+y+x-y}{2}\right) \cdot \cos \left(\frac{x+y-x+y}{2}\right)}{2 \cos \left(\frac{x+y+x-y}{2}\right) \cdot \sin \left(\frac{x+y-x+y}{2}\right)}=\frac{2 a}{2 b} \\ & {\left[\because \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2} \text { and } \sin x-\sin y=2 \cos \frac{x+y}{2} \cdot \sin \frac{x-y}{2}\right]} \\ & \Rightarrow \quad=\frac{\sin x \cdot \cos y}{\cos x \cdot \sin y}=\frac{a}{b} \\ & \Rightarrow \quad \frac{\tan x}{\tan y}=\frac{a}{b} \end{aligned}$$

Given that, $$\tan \theta = {{\sin \alpha - \cos \alpha } \over {\sin \alpha + \cos \alpha }}$$

$$ \Rightarrow \tan \theta = {{\cos \alpha (\tan \alpha - 1)} \over {\cos \alpha (\tan \alpha + 1)}}$$

$$ \Rightarrow \tan \theta = {{\tan \alpha - \tan {\pi \over 4}} \over {1 + \tan {\pi \over 4}\,.\,\tan \alpha }}\quad \left[\because{\tan {\pi \over 4} = 1} \right]$$

$$ \Rightarrow \tan \theta = \tan \left( {\alpha - {\pi \over 4}} \right)$$

$$ \Rightarrow \theta = \alpha - {\pi \over 4} \Rightarrow \alpha = \theta + {\pi \over 4}$$

$$\therefore \sin \alpha + \cos \alpha = \sin \left( {\theta + {\pi \over 4}} \right) + \cos \left( {\theta + {\pi \over 4}} \right)$$

$$ = \sin \theta \,.\,\cos {\pi \over 4} + \cos \theta \,.\,\sin {\pi \over 4} + \cos \theta \,.\,\cos {\pi \over 4} - \sin \theta \,.\,\sin {\pi \over 4}$$

$$= {1 \over {\sqrt 2 }}\sin \theta + {1 \over {\sqrt 2 }}\cos \theta + {1 \over {\sqrt 2 }}\cos \theta - {1 \over {\sqrt 2 }}\sin \theta \quad \left[\because{\sin {\pi \over 4} = \cos {\pi \over 4} = {1 \over {\sqrt 2 }}} \right]$$

$$ = {2 \over {\sqrt 2 }}\,.\,\cos \theta = \sqrt 2 \cos \theta $$

Given that, $\sin \theta+\cos \theta=1$

On squaring both sides, we get

$$\begin{aligned} & & \sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cdot \cos \theta & =1 \\ \Rightarrow & & 1+2 \sin \theta \cdot \cos & =1 \quad[\because \sin 2 x=2 \sin x \cos x]\\ \Rightarrow & & \sin 2 \theta & =0 \Rightarrow 2 \theta=n \pi+(-1)^n \cdot 0\\ \therefore & & \theta & =\frac{n \pi}{2} \end{aligned}$$

Alternate Method

$$\sin \theta + \cos \theta = 1$$

$$ \Rightarrow {1 \over {\sqrt 2 }}\,.\,\sin \theta + {1 \over {\sqrt 2 }}\,.\,\cos \theta = {1 \over {\sqrt 2 }}$$

$$ \Rightarrow \sin \theta \,.\,\cos {\pi \over 4} + \cos \theta \,.\,\sin {\pi \over 4} = {1 \over {\sqrt 2 }}\,\,\left[\because {\sin {\pi \over 4} = {1 \over {\sqrt 2 }} = \cos {\pi \over 4}} \right]$$

$$ \Rightarrow \sin \left( {\theta + {\pi \over 4}} \right) = \sin {\pi \over 4}\,\,[\because\sin (x + y) = \sin x\,.\,\cos y + \cos x\,.\,\sin y]$$

$$ \Rightarrow \theta + {\pi \over 4} = n\pi + {( - 1)^n}{\pi \over 4}$$

$$\therefore \quad \theta = n\pi + {( - 1)^n}{\pi \over 4} - {\pi \over 4}$$