Question

Vă rog tare frumos ajutați mă dau coroană!!!!!

Answer 1

$a)\:\frac{\sin x}{1-\cos x}=\frac{\sin x(1+\cos x)}{(1-\cos x)(1+\cos x)}=\frac{\sin x(1+\cos x)}{1-\cos^2x}=\frac{\sin x(1+\cos x)}{\sin^2x}=\frac{1+\cos x}{\sin x}$

$b)\:tg^2x-\sin^2x=\frac{\sin^2x}{\cos^2x}-\sin^2x=\frac{\sin^2x-\sin^x\cos^2x}{\cos^2x}=\frac{\sin^2x(1-\cos^2x)}{\cos^2x}=\\\\=\frac{\sin^2x}{\cos^2x}\cdor\sin^2x=tg^2x\cdot\sin^2x$

$c)\:1+tg^2x=1+\frac{\sin^2x}{\cos^2x}=\frac{\cos^2x+\sin^2x}{cos^2x}=\frac{1}{\cos^2x}$

$d)\:1+ctg^2x=1+\frac{\cos^2x}{\sin^2x}=\frac{sin^2x+\cos^2x}{\sin^2x}=\frac{1}{\sin^2x}$

$e)\:\frac{1}{\sin^2x}+\frac{1}{\cos^2x}=\frac{\cos^2x+\sin^2x}{\sin^2x\cos^2x}=\frac{1}{\sin^2x\cos^2x}=(\frac{1}{\sin x\cos x})^2=\\\\=(\frac{\sin^2x+\cos^2x}{\sin x\cos x})^2=(\frac{\sin^2x}{\sin x\cos x}+\frac{\cos^2x}{\sin x\cos x})^2=(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x})^2=\\\\=(tg\:x+ctg\:x)^2$