Question

Fie x masura unui unghi ascutit. Aratati ca:
a) tg²x-sin²x=tg²×sin²x
b)1+tg²x=$\frac{1}{cos ^{2}x }$
c)1+ctg²x=$\frac{1}{sin ^{2}x }$
d)$\frac{1}{sin ^{2}x }$ + $\frac{1}{cos ^{2}x }$ = (tg x+ ctg x)²
Toate subpunctele, va rog frumos!

Answer 1

[tex]a) tg^2 x-\sin ^2x=\dfrac{\sin ^2x}{\cos^2x}-\sin^2x=\dfrac{\sin^2x-\sin^2x\cdot \cos^2x}{\cos^2x}=\\ \\ =\dfrac{\sin^2x(1-\cos^2x)}{\cos^2x}=\dfrac{\sin^2x\cdot \sin^2x}{\cos^2x}=tg^2x\cdot \sin^2x\\ \\ b)tg^2x=\dfrac{\sin^2x}{\cos^2x}\\ tg^2x=\dfrac{1-\\cos^2x}{\cos^2x}\\ tg^2x=\dfrac{1}{\cos^2x}-1\\ 1+tg^2x=\dfrac{1}{\cos^2x}[/tex]
[tex]c) ctg^2x=\dfrac{\cos^2x}{\sin^2x}\\ ctg^2x=\dfrac{1-\sin^2x}{\sin^2x}\\ ctg^2x=\dfrac{1}{\sin^2x}-1\\ 1+ctg^2x=\dfrac{1}{\sin^2x}\\ \\ d)(tgx+ctgx)^2=tg^2x+2\cdot tgx\cdot ctg x+ctg^2x=\dfrac{\sin^2x}{\cos^2x}+2+\dfrac{cos^2x}{\sin^2x}=\\ =\dfrac{1-\cos^2x}{\cos^2x}+\dfrac{1-\sin^2x}{\sin^2x}+2=\dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x}-1-1+2=\\ =\dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x}[/tex]