Matemática
Matheuscarregal
4

O valor de (tg²x + 1).(sen²x - 1) é... PS: x pertence ao primeiro quadrante.

+0
(1) Respostas
Vanessa22

[latex]\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)=[/latex] [latex]\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}[/latex]

Adicionar resposta