If $\sec \theta = \frac{5}{3}$,then $\tan \theta = \ldots$

$\frac{1}{\tan ^{2} \theta}+1 = \dots$

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

Write 'True' or 'False' and justify your answer.
$(\tan \theta+2)(2 \tan \theta+1)=5 \tan \theta+\sec ^{2} \theta$

$2 \sin ^{2} \theta+4 \sec ^{2} \theta+5 \cot ^{2} \theta+2 \cos ^{2} \theta-4 \tan ^{2} \theta-5 \operatorname{cosec}^{2} \theta = \dots$

Simplify $(1+\tan ^{2} \theta)(1-\sin \theta)(1+\sin \theta)$