$\tan ^{2} \theta - \sec ^{2} \theta = \ldots \ldots \ldots$

Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

$\cos (40^{\circ}-\theta)-\sin (50^{\circ}+\theta) = \ldots \ldots \ldots \ldots$

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

State whether the following is 'True' or 'False' and justify your answer:
If $\cos A + \cos^2 A = 1$,then $\sin^2 A + \sin^4 A = 1$.

If $3 \theta$ is the measure of an acute angle and $\sin 3 \theta = \cos (\theta - 26^{\circ})$,then the value of $\theta$ is $\ldots \ldots \ldots \ldots$ (in $^{\circ}$)