$(\sin 80^{\circ} + \cos 10^{\circ})(\sin 80^{\circ} - \cos 10^{\circ}) = \ldots \ldots \ldots$

$\frac{\cos ^{2} 40^{\circ}+\cos ^{2} 50^{\circ}}{\sin ^{2} 40^{\circ}+\sin ^{2} 50^{\circ}}=\ldots \ldots \ldots \ldots$

The value of $\tan 15^{\circ}$ and $\ldots \ldots \ldots \ldots$ are equal.

If $\cos 9 \alpha = \sin \alpha$ and $9 \alpha < 90^{\circ},$ then the value of $\tan 5 \alpha$ is

$\tan (90^\circ - \theta) = \ldots \ldots \ldots$

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