$(1+\tan ^{2} \theta)(1-\cos ^{2} \theta) = \dots$

$\sin (45^{\circ}+\theta)-\cos (45^{\circ}-\theta)$ is equal to

$\tan (65^\circ - \theta) - \cot (25^\circ + \theta) - \sec (55^\circ - \theta) + \operatorname{cosec}(35^\circ + \theta) = \ldots \ldots \ldots \ldots$ (where,$0 < \theta < 25^\circ$)

If $\sin \theta + \cos \theta = p$ and $\sec \theta + \operatorname{cosec} \theta = q,$ then prove that $q(p^2 - 1) = 2p$.

Show that $\frac{\cos ^{2}\left(45^{\circ}+\theta\right)+\cos ^{2}\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \tan \left(30^{\circ}-\theta\right)}=1$

Show that $\tan ^{4} \theta+\tan ^{2} \theta=\sec ^{4} \theta-\sec ^{2} \theta$