Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$

[question] Question. Prove that $\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}=2 \sin x$ [/question] [solution] solution: It is known that $\sin A-\sin B=2 \cos \left(\frac{A+B}{2}\right) \sin \left(\frac{A-B}{2}\right), \cos ^{2} A-\sin ^{2} A=\cos 2 A$ $\therefore$ L.H.S. $=\frac{\sin x-\sin 3 x}{\sin ^{2} x-\cos ^{2} x}$ $=\frac{2 \cos \left(\frac{x+3 x}{2}\right) \sin \left(\frac{x-3 x}{2}\right)}{-\cos 2 x}$ $=\frac{2 \cos 2 x \sin (-x)}{-\cos 2 x}$ $=-2 \times(-\sin x)$ $=2 \sin x=$ R.H....

Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$

[question] Question. Prove that $\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}=\cot ^{2} x$ [/question] [solution] solution: L.H.S. $=\frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\frac{\pi}{2}+x\right)}$ $=\frac{[-\cos x][\cos x]}{(\sin x)(-\sin x)}$ $=\frac{-\cos ^{2} x}{-\sin ^{2} x}$ $=\cot ^{2} x$ $=$ R.HS. [/solution]...

$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$

[question] Question. $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$ [/question] [solution] solution: L.H.S. $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$ $=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$ $=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$ $=$ R.H.S. [/solution]...