We have x +  $\frac{1}{x}$ = $2cos\theta$,

Now $x^3$ +  $\frac{1}{x^3}$ = ${(x+\frac{1}{x})}^3$ - $3x\frac{1}{x}(x+\frac{1}{x})$

= $(2cos\theta {)}^3-3(2cos\theta )=8co{s}^3\theta -6cos\theta$

=$2(4co{s}^3\theta -3cos\theta )=2cos3\theta$.

Trick: Put x = 1 $\Rightarrow$ $\theta =0^{\circ }$.

Then $x^3$ +  $\frac{1}{x^3}$ = 2 = $2cos3\theta$.

We have $sin(\alpha -\beta )$ = $sin(\theta -\beta -\overline{\theta -\alpha })$

= $sin(\theta -\beta )cos(\theta -\alpha )-cos(\theta -\beta )sin(\theta -\alpha )$

= ba - $\sqrt{1-b^2}\sqrt{1-a^2}$

And $cos(\alpha -\beta )=cos(\theta -\beta -\overline{\theta -\alpha })$

= $cos(\theta -\beta )cos(\theta -\alpha )+sin(\theta -\beta )sin(\theta -\alpha )$

= $a\sqrt{1-b^2}+b\sqrt{1-a^2}$

∴ Given expression is $co{s}^2(\alpha -\beta )+2absin(\alpha -\beta )$

= $(a\sqrt{1-b^2}+b\sqrt{1-a^2}{)}^2$  + 2ab{$ab-\sqrt{1-a^2}\sqrt{1-b^2}$}

= $a^2+b^2$.

Trick:  Put $\alpha ={30}^{\circ }$, $\beta ={60}^{\circ }$ and $\theta ={90}^{\circ }$,

then a =  $\frac{1}{2}$, b =  $\frac{1}{2}$

∴ $co{s}^2(\alpha -\beta )+2absin(\alpha -\beta )$ =  $\frac{3}{4}$ +  $\frac{1}{2}$ × (- $\frac{1}{2}$) =  $\frac{1}{2}$

which is given by option (c).

Adding and subtracting the given relation,

we get $(m+n)=a{\cos }^3\alpha +3a\cos \alpha {\sin }^2\alpha$

     + $3a{\cos }^2\alpha .\sin \alpha +a{\sin }^3\alpha$

= $a(\cos \alpha +\sin \alpha {)}^3$

and similarly $(m-n)=a(\cos \alpha -\sin \alpha {)}^3$

Thus, $(m+n{)}^{\frac{2}{3}}+(m-n{)}^{\frac{2}{3}}$
$=a^{\frac{2}{3}}$ {$(\cos \alpha +\sin \alpha {)}^2+(\cos \alpha -\sin \alpha {)}^2$}
$=a^{\frac{2}{3}}$ $\left[2\right({\cos }^2\alpha +{\sin }^2\alpha \left)\right]$ = $2a^{\frac{2}{3}}$.

A.M. $\ge$  G.M.

$\Rightarrow$  $\frac{9ta{n}^2\theta +4co{t}^2\theta }{2}$ $\ge$ $\sqrt{4co{t}^2\theta .9ta{n}^2\theta }$

$\Rightarrow$  $9ta{n}^2\theta +4co{t}^2\theta$  $\ge$ 12

Therefore, the minimum value is 12.

Minimum value of $(3sin\theta +4cos\theta )$ is -$\sqrt{3^2+4^2}$ i.e., -5.

Given that sec$\theta$ =  $\frac{5}{4}$

sec$\theta$ =  $\frac{1+ta{n}^2\left(\frac{\theta }{2}\right)}{1-ta{n}^2\left(\frac{\theta }{2}\right)}$ $\Rightarrow$  $\frac{5}{4}$ =  $\frac{1+ta{n}^2\left(\frac{\theta }{2}\right)}{1-ta{n}^2\left(\frac{\theta }{2}\right)}$

$\Rightarrow$ $5-5ta{n}^2\left(\frac{\theta }{2}\right)=4+4ta{n}^2\left(\frac{\theta }{2}\right)$

$\Rightarrow$ $9ta{n}^2\left(\frac{\theta }{2}\right)$ = 1 $\Rightarrow$ $tan\left(\frac{\theta }{2}\right)$ =  $\frac{1}{3}$.

$\frac{co{t}^2{15}^{\circ }-1}{co{t}^2{15}^{\circ }+1}$ =  $\frac{\frac{co{s}^2{15}^{\circ }}{si{n}^2{15}^{\circ }}-1}{\frac{co{s}^2{15}^{\circ }}{si{n}^2{15}^{\circ }}+1}$

= $\frac{co{s}^2{15}^{\circ }-si{n}^2{15}^{\circ }}{co{s}^2{15}^{\circ }+si{n}^2{15}^{\circ }}$ = $cos\left({30}^{\circ }\right)$ =  $\frac{\sqrt{3}}{2}$

We have sin A =  $\frac{4}{5}$ and cos B = - $\frac{12}{13}$

Now, cos(A + B) = cos A cos B - sin A sin B

= $\sqrt{1-\frac{16}{25}}$$(-\frac{12}{13})$ -  $\frac{4}{5}$$(-\sqrt{1-\frac{144}{169}})$

= - $\frac{3}{5}$ × $\frac{12}{13}$ -  $\frac{4}{5}$$(-\frac{5}{13})$ = - $\frac{16}{65}$

(Since A lies in first quadrant and B lies in third quadrant).

$\frac{cotA}{1+cotA}$. $\frac{cotB}{1+cotB}$ =  $\frac{1}{(1+tanA)(1+tanB)}$

=  $\frac{1}{tanA+tanB+1+tanAtanB}$

                           [ ∵ tan(A + B) = $tan{225}^{\circ }$]

$\Rightarrow$ tanA + tan B = 1 - tan A tan B

=  $\frac{1}{1-tanAtanB+1+tanAtanB}$ =  $\frac{1}{2}$.

We have tan A = -
$\frac{1}{2}$ and tan B = -
$\frac{1}{3}$

Now, tan(A + B) = 
$\frac{tanA+tanB}{1-tanAtanB}$ = 
$\frac{-\frac{1}{2}-\frac{1}{3}}{1-\frac{1}{2}.\frac{1}{3}}$ = -1

$\Rightarrow$ tan(A + B) = tan
$\frac{3\pi }{4}$. Hence, A + B = 
$\frac{3\pi }{4}$.