Question

O valor de (tg²x + 1).(sen²x - 1) é...

PS: x pertence ao primeiro quadrante.

Answer 1

$\left(\tan^2x+1\right)\left(\sin^2x-1\right)=\\\\\\\left(\frac{\sin^2x}{\cos^2x}+1\right)\left(\sin^2x-1\right)=\\\\\\\left(\frac{\sin^2x+\cos^2x}{\cos^2x}\right)\left(\sin^2x-1\right)=\\\\\\\left(\frac{\overbrace{\sin^2x+\cos^2x}^{1}}{\cos^2x}\right)\left(\sin^2x-1\right)=$

$\left(\frac{1}{\cos^2x}\right)\left(\sin^2x-1\right)=\\\\\\\frac{\sin^2x}{\cos^2x}-\frac{1}{\cos^2x}=\\\\\\\boxed{\tan^2x-sec^2x}$

Answer 2

Resposta:

-1

Explicação passo-a-passo:

substitua o x por um valor do intervalo, como 0°:

(tg²0°+1) . (sen²0°-1) = (0+1) . (0-1) = -1