Question

Va rog sa ma ajutati !!!!!

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\it \dfrac{1+sin^2x}{2+ctg^2x}=\dfrac{1+sin^2x}{2+\dfrac{cos^2x}{sin^2x}}=\dfrac{1+sin^2x}{\dfrac{2sin^2x+cos^2x}{sin^2x}}=\dfrac{1+sin^2x}{\dfrac{sin^2x+sin^2x+cos^x}{sin^2x}}=\\ \\ \\ =\dfrac{1+sin^2x}{\dfrac{1+sin^x}{sin^2x}}=(1+sin^2x)\cdot\dfrac{sin^2x}{1+sin^2x}=sin^2x$

$\it Analog\ a\ doua\ frac\c{\it t}ie\ devine\ \ cos^2x,\ iar\ expresia\ se\ scrie: \\ \\ sin^2x+cos^2x=1\ (Adev\breve{a}rat)$