$$\begin{aligned} \text{Here,}\quad \frac{\sin 50 \Upsilon}{\sin 130 \Upsilon} & =\frac{\sin (180 \Upsilon-130 \Upsilon)}{\sin 130 \Upsilon} \\ & =\frac{\sin 130 \Upsilon}{\sin 130 \Upsilon}=1 \end{aligned} $$

Here,

$$\begin{aligned} &\begin{aligned} & k=\sin \left(\frac{\pi}{18}\right) \sin \left(\frac{5 \pi}{18}\right) \sin \left(\frac{7 \pi}{18}\right) \\ & =\sin 10 \Upsilon \sin50\Upsilon \sin70\Upsilon \\ & =\sin 10 \Upsilon \cos 40 r \cdot \cos 20 \Upsilon \\ & =\frac{1}{2} \sin 10 \Upsilon[2 \cos 40 \Upsilon \cdot \cos 20 \Upsilon] \\ & =\frac{1}{2} \sin 10 \Upsilon[\cos 60 \Upsilon+\cos 20 \Upsilon] \quad[\because 2 \cos x \cdot \cos y=\cos (x+y)+\cos (x-y)] \\ & =\frac{1}{2} \sin 10 r \cdot \frac{1}{2}+\frac{1}{2} \sin 10 \Upsilon \cos 20 \Upsilon \\ & =\frac{1}{4} \sin 10 \Upsilon+\frac{1}{4}[\sin 30 \Upsilon-\sin 10 \Upsilon] \\ & =\frac{1}{8} \end{aligned}\\ \end{aligned}$$

Given that,

$$\begin{aligned} \tan A & =\frac{1-\cos B}{\sin B} \\ & =\frac{1-1+2 \sin ^2 \frac{B}{2}}{2 \sin \frac{B}{2} \cdot \cos \frac{B}{2}}=\tan \frac{B}{2} \end{aligned}$$$$ \begin{array}{ll} \text { Now, } & \tan 2 A=\frac{2 \tan A}{1-\tan ^2 A} \\ \Rightarrow & \tan 2 A=\frac{2 \cdot \tan \frac{B}{2}}{1-\tan ^2 \frac{B}{2}} \\ \Rightarrow & \tan 2 A=\tan B \end{array}$$

Given that, $\sin x+\cos x=a$

On squaring both sides, we get

$$\begin{array}{rlrl} & (\sin x+\cos x)^2 =(a)^2 \\ \Rightarrow & \sin ^2 x+\cos ^2 x+2 \sin x \cos x =a^2 \\ \Rightarrow & \sin x \cdot \cos x =\frac{1}{2}\left(a^2-1\right) \end{array}$$

$$\begin{aligned} \text{(i) }\sin ^6 x+\cos ^6 x & =\left(\sin ^2 x\right)^3+\left(\cos ^2 x\right)^3 \\ & =\left(\sin ^2 x+\cos ^2 x\right)\left(\sin ^4 x-\sin ^2 x \cos ^2 x+\cos ^4 x\right) \\ & =\sin ^4 x+\cos ^4 x-\frac{1}{4}\left(a^2-1\right)^2 \\ & =\left(\sin ^2 x+\cos ^2 x\right)^2-2 \sin ^2 x \cos ^2 x-\frac{1}{4}\left(a^2-1\right)^2 \\ & =1-2 \cdot \frac{1}{4}\left(a^2-1\right)^2-\frac{1}{4}\left(a^2-1\right)^2=\frac{1}{4}\left[4-3\left(a^2-1\right)^2\right] \end{aligned}$$

$$\begin{aligned} \text{(ii) } |\sin x-\cos x| & =\sqrt{(\sin x-\cos x)^2} \\ & =\sqrt{\sin ^2 x+\cos ^2 x-2 \sin x \cos x} \\ & =\sqrt{1-2 \frac{1}{2}\left(a^2-1\right)}=\sqrt{1-a^2+1}=\sqrt{2-a^2} \end{aligned} $$

In right angled $\triangle A B C, \angle C=90 \Upsilon$

$$\begin{aligned} \therefore \quad \tan (A+B) & =\frac{\tan A+\tan B}{1-\tan A \tan B} \\ \Rightarrow \quad \frac{1}{0} & =\frac{\tan A+\tan B}{1-\tan A \tan B} \\ \Rightarrow \quad \tan A \tan B & =1 \quad \text{.... (i)}\\ \tan A+\tan B & =\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B} \\ & =\frac{\sin A}{\cos A}+\frac{\sin (90 \Upsilon-A)}{\cos (90 \Upsilon-A)} \\ & =\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} \\ & =\frac{\sin ^2 A+\cos A}{\sin A \cdot \cos A} \\ & =\frac{1}{\sin A \cdot \cos A}=\frac{2}{2 \cdot \sin A \cdot \cos A} \\ & =\frac{2}{\sin 2 A} \quad [\because \sin2x=2\sin x \cos x] \end{aligned}$$

So, the required equation is $x^2-\left(\frac{2}{\sin A}\right) x+1$