Show that $\left(\frac{1+{\tan }^2\mathrm{A}}{1+{\cot }^2\mathrm{A}}\right)={\left(\frac{1+\mathrm{tanA}}{1+\mathrm{cotA}}\right)}^2={\tan }^2\mathrm{A}$.

If $\cos \theta =\frac{2p}{p^2+1}$, $\left(p\ne \pm 1\right)$ then $\mathrm{cosec}\theta$ is equal to:

The value of $(\sin \theta + \cos \theta {)}^2=?$

1. $1+{\sin }^2\theta$
2. ${\sin }^2\theta +{\cos }^2\theta$ 
3.$1+2 \cos \theta \sin \theta$ 
4. ${\cos }^2 \theta + 1$

If $\sin \theta =\frac{1}{\sqrt{2}}$, then $(\tan \theta + \cos \theta )$ = 

1. $\frac{1}{\sqrt{2}}$
2. $\frac{2}{\sqrt{2}}$ 
3. $\frac{3}{\sqrt{2}}$
4. $\frac{(1+\sqrt{2})}{\sqrt{2}}$