$1+cos{56}^{\circ }+cos{58}^{\circ }-cos{66}^{\circ }=$
=$2co{s}^2{28}^{\circ }+2sin{62}^{\circ }.sin{4}^{\circ }$
=$2co{s}^2{28}^{\circ }+2cos{28}^{\circ }.sin{4}^{\circ }$
=$2cos{28}^{\circ }\left(cos{28}^{\circ }cos{86}^{\circ }\right)$
=$2cos{28}^{\circ }.2cos{57}^{\circ }.cos{29}^{\circ }$
=$4cos{28}^{\circ }cos{29}^{\circ }sin{33}^{\circ }$
Aliter: Apply the condition identify
$cosA+cosB-cosC=-1+4cos\frac{A}{2}cos\frac{B}{2}sin\frac{c}{2}$
$[\therefore {56}^{\circ }+{58}^{\circ }+{66}^{\circ }={180}^{\circ }]$
We get the value of required expression equal to $4cos{28}^{\circ }cos{29}^{\circ }sin{33}^{\circ }$.

$\sqrt{3}cosec{20}^{\circ }-sec{20}^{\circ }=\frac{\sqrt{3}}{sin{20}^{\circ }}-\frac{1}{cos{20}^{\circ }}$
$=\frac{\sqrt{3}cos{20}^{\circ }-sin{20}^{\circ }}{sin{20}^{\circ }cos{20}^{\circ }}=\frac{2[\frac{\sqrt{3}}{2}cos{20}^{\circ }-\frac{1}{2}sin{20}^{\circ }]}{\frac{2}{2}sin{20}^{\circ }cos{20}^{\circ }}$
$=\frac{4cos({20}^{\circ }+{30}^{\circ })}{sin{40}^{\circ }}=\frac{4cos{50}^{\circ }}{sin{40}^{\circ }}=\frac{4sin{40}^{\circ }}{sin{40}^{\circ }}=4$.

$sin{20}^{o}sin{40}^{\circ }sin{60}^{\circ }sin{80}^{\circ }$
=$\frac{1}{2}sin{20}^{\circ }sin{60}^{\circ }\left(2sin{40}^{\circ }sin{80}^{\circ }\right)$
=$\frac{1}{2}sin{20}^{\circ }sin{60}^{\circ }(cos{40}^{\circ }-cos{120}^{\circ })$
=$\frac{1}{2}.\frac{\sqrt{3}}{2}sin{20}^{\circ }(1-2si{n}^2{20}^{\circ }+\frac{1}{2})$
=$\frac{\sqrt{3}}{4}sin{20}^{\circ }(\frac{3}{2}-2si{n}^2{20}^{\circ })$
=$\frac{\sqrt{3}}{8}(3sin{20}^{\circ }-4si{n}^3{20}^{\circ })$
=$\frac{\sqrt{3}}{8}sin{60}^{\circ }=\frac{\sqrt{3}}{8}.\frac{\sqrt{3}}{2}=\frac{3}{16}$

$\left(\frac{sin2A}{1+cos2A}\right)$$\left(\frac{cosA}{1+cosA}\right)$

=
$\frac{2sinAcosA}{2co{s}^{2}A}$ 
$\frac{cosA}{1+cosA}$ = 
$\frac{sinA}{1+cosA}$ = tan 
$\frac{A}{2}$

Given that tan $\theta$ =  $\frac{b}{a}$.

Now, a cos 2$\theta$ + b sin 2 $\theta$ = a$\left(\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }\right)$ + b$\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)$

Putting $tan\theta$ =  $\frac{b}{a}$, we get

= a$\left(\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}\right)$ + b$\left(\frac{2\frac{b}{a}}{1+\frac{b^2}{a^2}}\right)$ = a$\left(\frac{a^2-b^2}{a^2+b^2}\right)$ + b$\left(\frac{2ba}{a^2+b^2}\right)$

=  $\frac{1}{(a^2+b^2)}$ $a^3-a{b}^2+2a{b}^2$ =  $\frac{a(a^2+b^2)}{a^2+b^2}$ = a.

$co{s}^2(\frac{\pi }{3}-x)$ - $co{s}^2(\frac{\pi }{3}+x)$

= $\{cos(\frac{\pi }{3}-x)+cos(\frac{\pi }{3}+x)\}$  $\{cos(\frac{\pi }{3}-x)-cos(\frac{\pi }{3}+x)\}$

= $\left\{2cos\frac{\pi }{3}cosx\right\}$ $\left\{2sin\frac{\pi }{3}sinx\right\}$

= sin $\frac{2\pi }{3}$ sin 2x =  $\frac{\sqrt{3}}{2}$ sin 2x

Its maximum value is  $\frac{\sqrt{3}}{2}$, {- 1 $\le$ sin 2x $\le$ 1}.

Given that tan 
$\frac{A}{2}$ = 
$\frac{3}{2}$.

$\frac{1+cosA}{1-cosA}$ = 
$\frac{2co{s}^2\frac{A}{2}}{2si{n}^2\frac{A}{2}}$ = $co{t}^2\frac{A}{2}$ = ${\left(\frac{2}{3}\right)}^2$ = 
$\frac{4}{9}$

We have cos A = 
$\frac{\sqrt{3}}{2}$ $\Rightarrow$ A = ${30}^{\circ }$

$\Rightarrow$ tan 3A = $tan{90}^{\circ }$ = $\infty$.

$\frac{1}{4}$$\sqrt{3}cos{23}^{\circ }-sin{23}^{\circ }$

 = $\frac{1}{2}$$cos{30}^{\circ }cos{23}^{\circ }-sin{30}^{\circ }sin{23}^{\circ }$

=  $\frac{1}{2}$ $cos({30}^{\circ }+{23}^{\circ })$ =  $\frac{1}{2}$ $cos{53}^{\circ }$.

Given expression is

${\sin }^2{5}^{\circ }+{\sin }^2{10}^{\circ }+{\sin }^2{15}^{\circ }+..............+{\sin }^2{85}^{\circ }+{\sin }^2{90}^{\circ }$.

We know that $\sin {90}^{\circ }=1\text{or}{\sin }^2{90}^{\circ }=1$

Similarly, $\sin {45}^{\circ }=$$\frac{1}{\sqrt{2}}$ or ${\sin }^2{45}^{\circ }=\frac{1}{2}$ and the angles are in A.P. of 18 terms. We also know that

${\sin }^2{85}^{\circ }=\left[\sin \right({90}^{\circ }-5^{\circ }){]}^2={\cos }^2{5}^{\circ }$.

Therefore from the complementary rule, we find ${\sin }^2{5}^{\circ }+{\sin }^2{85}^{\circ }={\sin }^2{5}^{\circ }+{\cos }^2{5}^{\circ }=1$

Therefore,

${\sin }^2{5}^{\circ }+{\sin }^2{10}^{\circ }+{\sin }^2{15}^{\circ }+.............+{\sin }^2{85}^{\circ }+{\sin }^2{90}^{\circ }$

$=(1+1+1+1+1+1+1+1)+1+\frac{1}{2}=9\frac{1}{2}$.