$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

Prove the following identities,where the angles involved are acute angles for which the expressions are defined:
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$

State whether the following is true or false. Justify your answer.
The value of $\sin \theta$ increases as $\theta$ increases.

If $\sin ( A - B ) = \frac{1}{2}$,$\cos ( A + B ) = \frac{1}{2}$,$0^{\circ} < A + B \leq 90^{\circ}$,and $A > B$,find $A$ and $B$.

Evaluate:
$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$

Show that:
$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} = 1$
$(ii)$ $\cos 38^{\circ} \cos 52^{\circ} - \sin 38^{\circ} \sin 52^{\circ} = 0$