We know that,$1+ta{n}^2\theta =se{c}^2\theta \text{and}1+co{t}^2\theta =cose{c}^2\theta$
$\frac{1}{1+ta{n}^2\theta }+\frac{1}{1+co{t}^2\theta }=\frac{1}{se{c}^2\theta }+\frac{1}{cose{c}^2\theta }$
$=co{s}^2\theta +si{n}^2\theta$

$=1$

$\Rightarrow (\sin A+\text{cosec A}{)}^2-(\sin A-\text{cosec A}{)}^2$

$=4\sin A\text{cosec A}$ 
as $\left[\right(a+b{)}^2-(a-b{)}^2=4ab]$

now, $(\sin A\times \text{cosec A}=\sin A\times \frac{1}{\sin A}=1)$
$\therefore 4.\sin A\text{cosec A}=4\times 1=4$

$\text{We know that}tan{30}^{\circ }=\frac{1}{\sqrt{3}}.$

$\Rightarrow \frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}=\frac{1-(\frac{1}{\sqrt{3}}{)}^2}{1+(\frac{1}{\sqrt{3}}{)}^2}$

$=\frac{1-\left(\frac{1}{3}\right)}{1+\left(\frac{1}{3}\right)}$

$=\frac{3-1}{3+1}$

$=\frac{2}{4}$ = $\frac{1}{2}=0.5$ 

The given expression is 
$(\frac{1+\tan \theta }{1+\cot \theta }{)}^2=(\frac{1+\tan \theta }{1+\frac{1}{\tan \theta }}{)}^2$
                 
$=(\frac{1+\tan \theta }{\frac{\tan \theta +1}{\tan \theta }}{)}^2$
                 
$=(\frac{\left(\tan \theta \right)(1+\tan \theta )}{\tan \theta +1}{)}^2$
               
$={\tan }^2\theta$

$\frac{3\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }$ = $\frac{5}{\sin \theta }$
$\frac{3{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{5}{\sin \theta }$
$3{\sin }^2\theta +{\cos }^2\theta =5\cos \theta$
3(1 - ${\cos }^2\theta$) + ${\cos }^2\theta$ = 5 $\cos \theta$
2${\cos }^2\theta$ + 5$\cos \theta$ - 3 = 0 
2$\cos \theta$  [$\cos \theta$ + 3] - 1($\cos \theta$ + 3) = 0
($\cos \theta$ + 3) (2$cos\theta$ - 1) = 0
$\cos \theta$ = -3     or     $\cos \theta$ = $\frac{1}{2}$
Note that $\cos \theta$ = -3  is not possible as $-1\le \cos \theta \le 1$                     
 Thus, $\theta$ = ${60}^{\circ }$

$ta{n}^{2}A+co{t}^{2}A+2\text{can be written as}(1+ta{n}^{2}A)+(1+co{t}^{2}A)$
We know that, $1+ta{n}^{2}A=se{c}^{2}A\text{and}1+co{t}^{2}A=cose{c}^{2}A$
$\Rightarrow ta{n}^{2}A+co{t}^{2}A+2=se{c}^{2}A+cose{c}^{2}A$

$=\frac{1}{co{s}^{2}A}+\frac{1}{si{n}^{2}A}$

$=\frac{si{n}^{2}A+co{s}^{2}A}{co{s}^{2}Asi{n}^{2}A}...(si{n}^{2}A+co{s}^{2}A=1)$

$=\frac{1}{co{s}^{2}Asi{n}^{2}A}$

$=se{c}^{2}Acose{c}^{2}A$

We know that, ${\sin }^{2}A+{\cos }^{2}A=1.$
 $1){\cos }^{2}A-{\sin }^{2}A=(1-{\sin }^{2}A)-{\sin }^{2}A$
$=1-2{\sin }^{2}A$
$2){\cos }^{2}A-{\sin }^{2}A={\cos }^{2}A-(1-{\cos }^{2}A)$
$={\cos }^{2}A-1+{\cos }^{2}A$
$=2{\cos }^{2}A-1$

$A+B+C={180}^{\circ }$
(sum of angles of a triangle)
$⟹B+C={180}^{\circ }-A$
$\therefore \frac{B+C}{2}=\frac{{180}^{\circ }-A}{2}$
$\frac{B+C}{2}={90}^{\circ }-\frac{A}{2}$
Taking sine on both sides we get,
$\sin \left(\frac{B+C}{2}\right)=\sin ({90}^{\circ }-\frac{A}{2})$
$\therefore \sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}$

$\frac{sin\theta +1-cos\theta }{cos\theta -1+sin\theta }$
Dividing the numerator and denominator by $cos\theta$ we get,
$\frac{\frac{sin\theta }{cos\theta }+\frac{1}{cos\theta }-\frac{cos\theta }{cos\theta }}{\frac{cos\theta }{cos\theta }-\frac{1}{cos\theta }+\frac{sin\theta }{cos\theta }}$
$=\frac{tan\theta +sec\theta -1}{1-sec\theta +tan\theta }$
$\Rightarrow \frac{(tan\theta +sec\theta )-(se{c}^2\theta -ta{n}^2\theta )}{1-sec\theta +tan\theta }....(1+{\tan }^2\theta ={\sec }^2\theta )$
$=\frac{(\tan \theta +\sec \theta )-(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )}{1-\sec \theta +\tan \theta }$
$=\frac{(\tan \theta +\sec \theta )(1-\sec \theta +\tan \theta )}{(1-\sec \theta +\tan \theta )}$
$=\tan \theta +\sec \theta$
$=\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }$
$=\frac{\sin \theta +1}{\cos \theta }$
[Multiplying the numerator and denominator by $(1-\sin \theta )$]
$=\frac{1+\sin \theta }{\cos \theta }\times \frac{1-\sin \theta }{1-\sin \theta }$
$=\frac{1-{\sin }^2\theta }{\cos \theta (1-\sin \theta )}$
$=\frac{{\cos }^2\theta }{\cos \theta (1-\sin \theta )}$
$=\frac{\cos \theta }{1-\sin \theta }$

$\frac{\sin A-2{\sin }^{3}A}{2{\cos }^{3}A-cosA}=\frac{\sin A(1-2{\sin }^{2}A)}{\cos A(2{\cos }^{2}A-1)}$
$=\frac{\sin A({\sin }^{2}A+{\cos }^{2}A-2{\sin }^{2}A)}{\cos A(2{\cos }^{2}A-({\sin }^{2}A+{\cos }^{2}A\left)\right)}$
$=\frac{\sin A({\cos }^{2}A-{\sin }^{2}A)}{\cos A({\cos }^{2}A-{\sin }^{2}A)}$
$=\tan A$