The equation ${(a + b)^2} = 4ab\,{\sin ^2}\theta $ is possible only when

Prove that $\cos \left(\frac{3 \pi}{4}+x\right)-\cos \left(\frac{3 \pi}{4}-x\right)=-\sqrt{2} \sin x$

If ${\tan ^2}\alpha {\tan ^2}\beta + {\tan ^2}\beta {\tan ^2}\gamma + {\tan ^2}\gamma {\tan ^2}\alpha $$ + 2{\tan ^2}\alpha {\tan ^2}\beta {\tan ^2}\gamma = 1,$ then the value of ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $ is

If $x + \frac{1}{x} = 2\cos \alpha $, then ${x^n} + \frac{1}{{{x^n}}} = $

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ ........\tan 89^\circ = $

If $\sec \theta + \tan \theta = p,$ then $\tan \theta $ is equal to