Let $f\left(\theta \right)$ = $5si{n}^2\theta +4co{s}^2\theta$ = $4+si{n}^2\theta$

∴ $f\left(\theta \right)$ $\ge$ 4 + 0                         ($\because si{n}^2\theta \ge 0$) 

∴  The minimum value of $f\left(\theta \right)$ is 4.

Let 3 = $rcos\alpha$, 4 = $rsin\alpha$, so r = 5

$f\left(\theta \right)$ = $r.(cos\alpha cos\theta +sin\alpha sin\theta )$ =  $5.cos(\theta -\alpha )$

∴ The maximum value of $f\left(\theta \right)$ = 5.1 = 5.

{Since the maximum value of $cos(\theta -\alpha )$ = 1}.

Aliter:  As we know that, the maximum value of $asin\theta +bcos\theta$ is +$\sqrt{a^2+b^2}$ and the minimum value

is -$\sqrt{a^2+b^2}$. Therefore, the maximum value is $(3cos\theta +4sin\theta )$ = + $\sqrt{3^2+(-4{)}^2}$ = 5 and the minimum value is -5.

a cos $\theta$ + b sin $\theta$ = $\sqrt{a^2+b^2}$ $(\frac{acos\theta }{\sqrt{a^2+b^2}}+\frac{bsin\theta }{\sqrt{a^2+b^2}})$

                                                  =  $\sqrt{a^2+b^2}$ $sin(\theta +\varphi )$

Since, $-1\le sin(\theta +\varphi )$ $\le$ 1,

Then -$\sqrt{a^2+b^2}$ $\le$  $sin(\theta +\varphi )$ $\le$  $\sqrt{a^2+b^2}$.

Given that $cos3\theta =\alpha cos\theta +\beta co{s}^3\theta$

But $cos3\theta =4co{s}^3\theta -3cos\theta$ $\Rightarrow$ $(\alpha ,\beta )$ = (-3, 4).

We have sin A = sin B and cos A = cos B

$2cos\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)=0,-2sin\left(\frac{A+B}{2}\right)sin\left(\frac{A-B}{2}\right)=0$

Hence, sin$\left(\frac{A-B}{2}\right)$ = 0.

$\frac{1}{tan3A-tanA}$ -  $\frac{1}{cot3A-cotA}$ 

= $\frac{1}{tan3A-tanA}$ +  $\frac{tanAtan3A}{tan3A-tanA}$ = $\frac{1}{tan2A}$ = cot 2A.

We know that sin(A + B) = sinA cosB + cosA sinB

=  $\frac{1}{\sqrt{10}}$$\sqrt{1-\frac{1}{5}}$ + $\frac{1}{\sqrt{5}}$$\sqrt{1-\frac{1}{10}}$

= $\frac{1}{\sqrt{10}}$$\sqrt{\frac{4}{5}}$ + $\frac{1}{\sqrt{5}}$$\sqrt{\frac{9}{10}}$ =  $\frac{1}{\sqrt{50}}$(2 + 3) =  $\frac{5}{\sqrt{50}}$ =  $\frac{1}{\sqrt{2}}$

$\Rightarrow$ sin(A + B) = sin $\frac{\pi }{4}$

Hence, A + B = $\frac{\pi }{4}$.

cosec A - 2 cot 2A cos A = 
$\frac{1}{sinA}$ - 
$\frac{2cosAcos2A}{sin2A}$

= 
$\frac{1}{sinA}$ - 
$\frac{2cosAcos2A}{2sinAcosA}$ = 
$\frac{1-cos2A}{sinA}$ = 
$\frac{2si{n}^{2}A}{sinA}$

= 2sin A.

$(cos\alpha +cos\beta {)}^2+(sin\alpha +sin\beta {)}^2$

= $co{s}^2\alpha +co{s}^2\beta +2cos\alpha cos\beta +si{n}^2\alpha +si{n}^2\beta +2sin\alpha sin\beta$

= $2\left(1+cos(\alpha -\beta )\right)$  = $4co{s}^2\left(\frac{\alpha -\beta }{2}\right)$.

2 tan(A - B) = 2$\left(\frac{tanA-tanB}{1+tanAtanB}\right)$

= 2 
$\frac{(2tanB+cotB-tanB)}{1+(2tanB+cotB)tanB}$ = 2
$\frac{tanB+cotB}{2(1+ta{n}^{2}B)}$

= 
$\frac{cotB(ta{n}^{2}B+1)}{(1+ta{n}^{2}B)}$ = cot B.