$\begin{aligned} \text { LHS } & =\frac{\tan A+\sec A-1}{\tan A-\sec A+1} \\ & =\frac{\tan A+\sec A-\left(\sec ^2 A-\tan ^2 A\right)}{(\tan A-\sec A+1)} \quad\left[\because \sec ^2 A-\tan ^2 A=1\right] \\ & =\frac{(\tan A+\sec A)-(\sec A+\tan A)(\sec A-\tan A)}{(1-\sec A+\tan A)} \\ & =\frac{(\sec A+\tan A)(1-\sec A+\tan A)}{1-\sec A+\tan A} \\ & =\sec A+\tan A=\frac{1}{\cos A}+\frac{\sin A}{\cos A} \quad \quad \text { Hence proved. } \\ & =\frac{1+\sin A}{\cos A}=\text { RHS } \quad \text {Hence proved.}\end{aligned}$

$\begin{aligned} & \text { Given that, } \frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y \\ & \text { Now, } \quad \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\frac{(1-\cos \alpha+\sin \alpha)}{(1+\sin \alpha)} \cdot \frac{(1+\cos \alpha+\sin \alpha)}{(1+\cos \alpha+\sin \alpha)}\end{aligned}$

$\begin{aligned} & =\frac{\{(1+\sin \alpha)-\cos \alpha\}}{(1+\sin \alpha)} \cdot \frac{\{(1+\sin \alpha)+\cos \alpha\}}{(1+\cos \alpha+\sin \alpha)} \\ & =\frac{(1+\sin \alpha)^2-\cos ^2 \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{\left(1+\sin ^2 \alpha+2 \sin \alpha\right)-\cos ^2 \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)}\end{aligned}$

$\begin{aligned} & =\frac{1+\sin ^2 \alpha+2 \sin \alpha-1+\sin ^2 \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{2 \sin ^2 \alpha+2 \sin \alpha}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{2 \sin \alpha(1+\sin \alpha)}{(1+\sin \alpha)(1+\sin \alpha+\cos \alpha)} \\ & =\frac{2 \sin \alpha}{1+\sin \alpha+\cos \alpha}=y\quad \text{Hence proved.}\end{aligned}$

Given that,

$$\begin{aligned} & m \sin \theta=n \sin (\theta+2 \alpha) \\ & \therefore \quad \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{m}{n} \end{aligned}$$

Using componendo and dividendo, we get

$$\begin{gathered} \frac{\sin (\theta+2 \alpha)+\sin \theta}{\sin (\theta+2 \alpha)-\sin \theta}=\frac{m+n}{m-n} \\ \Rightarrow \quad \frac{2 \sin \left(\frac{\theta+2 \alpha+\theta}{2}\right) \cdot \cos \left(\frac{\theta+2 \alpha-\theta}{2}\right)}{2 \cos \left(\frac{\theta+2 \alpha+\theta}{2}\right) \cdot \sin \left(\frac{\theta+2 \alpha-\theta}{2}\right)}=\frac{m+n}{m-n} \\ \Rightarrow \quad [\because \sin x+\sin y=2 \sin \frac{x+y}{2} \cdot \cos \frac{x-y}{2} \text { and } \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \cdot \frac{x-y}{2}] \\ \Rightarrow \quad \frac{\sin (\theta+\alpha) \cdot \cos \alpha}{\cos (\theta+\alpha) \cdot \sin \alpha}=\frac{m+n}{m-n} \\ \Rightarrow \quad \tan (\theta+\alpha) \cdot \cot \alpha=\frac{m+n}{m-n} \quad \text { Hence proved.} \end{gathered}$$

Given that, $$\cos (\alpha + \beta ) = {4 \over 5}$$ and $$\sin (\alpha - \beta ) = {5 \over {13}}$$

$$ \Rightarrow \sin (\alpha + \beta ) = \sqrt {1 - {{16} \over {25}}} = \sqrt {{9 \over {25}}} = \pm {3 \over 5}$$

$\therefore$ $$\sin (\alpha + \beta ) = {3 \over 5}$$

and $$\cos (\alpha - \beta ) = \sqrt {1 - {{25} \over {169}}} = \sqrt {{{144} \over {169}}} = \pm {{12} \over {13}}$$

$\therefore$ $$\cos (\alpha - \beta ) = {{12} \over {13}}$$

Now, $$\tan (\alpha + \beta ) = {{\sin (\alpha + \beta )} \over {\cos (\alpha + \beta )}}\quad$$ [since, $\alpha$ lies between 0 and ${\pi \over 4}$]

$$ = {{{3 \over 5}} \over {{4 \over 5}}} = {3 \over 4}$$

and $$\tan (\alpha - \beta ) = {{\sin (\alpha - \beta )} \over {\cos (\alpha - \beta )}} = {{{5 \over {13}}} \over {{{12} \over {13}}}} = {5 \over {12}}$$

$$\therefore \quad \tan 2\alpha = \tan (\alpha + \beta + \alpha - \beta )$$

$$ = {{\tan (\alpha + \beta ) + \tan (\alpha - \beta )} \over {1 - \tan (\alpha + \beta )\,.\,\tan (\alpha - \beta )}}\quad$$ $$\left[ \because {\tan (x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x\,.\,\tan y}}} \right]$$

$$ = {{{3 \over 4} + {5 \over {12}}} \over {1 - {3 \over 4}\,.\,{5 \over {12}}}} = {{{{9 + 15} \over {12}}} \over {{{16 - 15} \over {16}}}} = {{12 \times 16} \over {12 \times 11}} = {{56} \over {33}}$$

Given that, $\tan x=\frac{b}{a}$

$\begin{aligned} \therefore \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}} & =\frac{\sqrt{(a+b)^2}+\sqrt{(a-b)^2}}{\sqrt{(a-b)(a+b)}} \\ & =\frac{(a+b)+(a-b)}{\sqrt{a^2-b^2}}=\frac{2 a}{\sqrt{a^2-b^2}}=\frac{2 a}{a \sqrt{1-\left(\frac{b}{a}\right)^2}} \quad\left[\because \frac{b}{a}=\tan x\right] \\ & =\frac{2}{\sqrt{1-\tan ^2 x}}=\frac{2 \cos x}{\sqrt{\cos ^2 x-\sin ^2 x}} \quad\left[\because \cos 2 x=\cos ^2 x-\sin ^2 x\right] \\ & =\frac{2 \cos x}{\sqrt{\cos 2 x}}\end{aligned}$