(1)

$\sin x=\frac{-3}{5},$$\pi <x<\frac{3\pi }{2}$

$\cos x=\sqrt{1-{\sin }^{2}x}=\sqrt{1-\frac{9}{25}}=\frac{-4}{5}$                 $[$$\because$ $x$ lies in third quadrant$]$

Now,  $\tan x=\frac{3}{4}$

So, $80({\tan }^{2}x-\cos x)=80(\frac{9}{16}+\frac{4}{5})=45+64=109$

(1)

$\frac{3\cos {36}^o+5\sin {18}^o}{5\cos {36}^o-3\sin {18}^o}$$=\frac{3\left[{\displaystyle \frac{1+\sqrt{5}}{4}}\right]+5\left[{\displaystyle \frac{\sqrt{5}-1}{4}}\right]}{5\left[{\displaystyle \frac{1+\sqrt{5}}{4}}\right]-3\left[{\displaystyle \frac{\sqrt{5}-1}{4}}\right]}$                                     $[\because \sin {18}^o=\frac{\sqrt{5}-1}{4} \mathrm{and} \cos {36}^o=\frac{1+\sqrt{5}}{4}]$

      $=\frac{3+3\sqrt{5}+5\sqrt{5}-5}{5+5\sqrt{5}-3\sqrt{5}+3}=\frac{8\sqrt{5}-2}{2\sqrt{5}+8}=\frac{4\sqrt{5}-1}{4+\sqrt{5}}$

      $=\frac{(4\sqrt{5}-1)(4-\sqrt{5})}{(\sqrt{5}+4)(4-\sqrt{5})}=\frac{16\sqrt{5}-20-4+\sqrt{5}}{16-5}$

      $=\frac{17\sqrt{5}-24}{11}=\frac{a\sqrt{5}-b}{c}$                              (Given)

$\therefore$   $a=17,$$b=24,$$c=11$

$\therefore$  $a+b+c=17+24+11=52$

(3)

As $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$

$=\frac{{\displaystyle \frac{1}{\sqrt{x(x^2+x+1)}}}+{\displaystyle \frac{\sqrt{x}}{\sqrt{x^2+x+1}}}}{1-{\displaystyle \frac{1}{x^2+x+1}}}$

$=\frac{1+x}{\sqrt{x(x^2+x+1)}}\times \frac{x^2+x+1}{x^2+x}=\frac{1+x}{\sqrt{x}}\times \frac{\sqrt{x^2+x+1}}{x(1+x)}$

$=\sqrt{\frac{x^2+x+1}{x^3}}=\sqrt{x^{-1}+x^{-2}+x^{-3}}=\tan C$

$\Rightarrow A+B=C$

(3)

We have, 

$\frac{2\cos \theta +i\sin \theta }{\cos \theta –3i\sin \theta }=\frac{2\cos \theta +i\sin \theta }{\cos \theta –3i\sin \theta }\times \frac{\cos \theta +3i\sin \theta }{\cos \theta +3i\sin \theta }$

                              $=\frac{2{\cos }^2\theta +6i\cos \theta \sin \theta +i\sin \theta \cos \theta –3{\sin }^2\theta }{{\cos }^2\theta +9{\sin }^2\theta }$

                               $=\frac{2{\cos }^2\theta –3{\sin }^2\theta }{{\cos }^2\theta +9{\sin }^2\theta }+\frac{i\left(7\cos \theta \sin \theta \right)}{{\cos }^2\theta +9{\sin }^2\theta }$

Now, $1+10\mathrm{Re}\left(\frac{2 \cos \theta +i \sin \theta }{\cos \theta –3i \sin \theta }\right)=0$

$\Rightarrow 1+\frac{20{\cos }^2\theta –30{\sin }^2\theta }{{\cos }^2\theta +9{\sin }^2\theta }=0$

$\Rightarrow {\cos }^2\theta ={\sin }^2\theta$

$\Rightarrow {\tan }^2\theta =1 \Rightarrow \tan \theta =\pm 1$

$\Rightarrow \theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$

$\therefore \sum _{\theta \in A}{\theta }^2=\frac{{\pi }^2}{16}+\frac{9{\pi }^2}{16}+\frac{25{\pi }^2}{16}+\frac{49{\pi }^2}{16}=\frac{84{\pi }^2}{16}=\frac{21{\pi }^2}{4}$

(2)

We have, 

$\sin 70°\left(\frac{\cos 10°\cos 70°–\sin 10°\sin 70°}{\sin 10° \sin 70°}\right)$

$=\sin 70°\left[\frac{\cos 80°}{\sin 10° \sin 70°}\right]$          [$\because$ cos (A + B) = cos A cos B – sin A sin B]

$=\frac{\cos 80°}{\cos 80°}=1$        [$\because \sin (90°–\theta )=\cos \theta$]

(4)

The given expression can be written as,

$\frac{1}{\sin {\displaystyle \frac{\pi }{6}}}\sum _{r=1}^{13}\frac{\sin \left[\right({\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{r\pi }{6}})–[\left({\displaystyle \frac{\pi }{4}}\right)+(r–1){\displaystyle \frac{\pi }{6}}\left]\right]}{\sin ({\displaystyle \frac{\pi }{4}}+(r–1\left){\displaystyle \frac{\pi }{6}}\right) \sin ({\displaystyle \frac{\pi }{4}}+{\displaystyle \frac{r\pi }{6}})}$

$=\frac{1}{\sin {\displaystyle \frac{\pi }{6}}}\sum _{r=1}^{13}\left(\cot \right(\frac{\pi }{4}+(r–1)\frac{\pi }{6})–\cot (\frac{\pi }{4}+\frac{\mathrm{r\pi }}{6}\left)\right)$

                    $[\because \frac{\sin (A–B)}{\sin A \sin B}=\cot B–\cot A]$

$=2[\cot \frac{\pi }{4}–\cot (\frac{\mathrm{\pi }}{4}+\frac{13\mathrm{\pi }}{6}\left)\right]=2(1–2+\sqrt{3})$

$=2\sqrt{3}–2=a\sqrt{3}+b$          [Given]

$\therefore a=2, b=–2$

So, $a^2+b^2=8$

(4)

Area of $△OAB=\frac{1}{2}\times 1\times 1=\frac{1}{2}$

Now, Area of $△$AMN = $\frac{4}{9}$ Area of $△$OAB

$=\frac{4}{9}\times \frac{1}{2}=\frac{2}{9}$          ... (i)

Since, $\angle$OAB = 45°, then let $\angle$MAN = $\theta$ and $\angle$MAO = 45° – $\theta$, then AM = sec (45° – $\theta$);

$AN=AM\times \frac{AN}{AM}=\sec (45°–\theta ) \cos \theta$

and $MN=AM\times \frac{MN}{AM}=\sec (45°–\theta ) \sin \theta$

Now, $\mathrm{Area} \mathrm{of} △AMN=\frac{1}{2}\times AN\times MN$

                                             $=\frac{1}{2}\times {\sec }^2(45°–\theta )\sin \theta \cos \theta =\frac{2}{9} [\mathrm{From} (i\left)\right]$

$\Rightarrow \frac{\sin 2\theta }{1+\sin 2\theta }=\frac{4}{9} \Rightarrow \sin 2\theta =\frac{4}{5}$

$\Rightarrow 2{\tan }^2\theta –5\tan \theta +2=0$

$\Rightarrow \tan \theta =\frac{1}{2}, 2 (\mathrm{Rejected} \mathrm{as} \theta <45°)$

$\because \angle OBA=45°$

$\therefore \tan 45°=\frac{MN}{BN} \Rightarrow MN=BN$

Now, $\mathrm{In} △AMN, \cot \theta =\frac{AN}{MN}=\frac{AN}{BN}=\frac{\lambda }{1} \Rightarrow \lambda =2$      $[\because \tan \theta =\frac{1}{2}]$

Hence, required sum is 2.

(1)

We have, $\sin x+{\sin }^{2}x=1$

$\Rightarrow \sin x=1–{\sin }^{2}x \Rightarrow \sin x={\cos }^{2}x \Rightarrow \tan x=\cos x$

Now, $({\cos }^{12}x+{\tan }^{12}x)+3({\cos }^{10}x+{\tan }^{10}x+{\cos }^{8}x+{\tan }^{8}x)+({\cos }^{6}x+{\tan }^{6}x)$

$=({\cos }^{12}x+{\cos }^{12}x)+3({\cos }^{10}x+{\cos }^{10}x+{\cos }^{8}x+{\cos }^{8}x)+({\cos }^{6}x+{\cos }^{6}x)$

$=2{\cos }^{12}x+6{\cos }^{10}x+6{\cos }^{8}x+2{\cos }^{6}x$

$=2{\sin }^{6}x+6{\sin }^{5}x+6{\sin }^{4}x+2{\sin }^{3}x$          [$\because \sin x={\cos }^{2}x$]

$=2{\sin }^{3}x[{\sin }^{3}x+3{\sin }^{2}x+3\sin x+1]$

$=2{\sin }^{3}x\left[{(1+\sin x)}^3\right]$

$=2\left[{(\sin x+{\sin }^{2}x)}^3\right]=2$          {$\because \sin x+{\sin }^{2}x=1$}