Question

Tenho dúvidas na alinea b poderiam resolver a ?

Answer 1

$\mathrm{tg\,}x\left(\cos x+\dfrac{1-\mathrm{sen^{2}\,}x}{\mathrm{sen\,}x} \right )\\ \\ \\ =\mathrm{tg\,}x\left(\cos x+\dfrac{\cos^{2} x}{\mathrm{sen\,}x} \right )\\ \\ \\ =\mathrm{tg\,}x \left(\cos x+\cos x\cdot \dfrac{\cos x}{\mathrm{sen\,}x} \right )$


Colocando o $\cos x$ em evidência, temos

$=\mathrm{tg\,}x \cdot \cos x\left(1+\dfrac{\cos x}{\mathrm{sen\,}x} \right )\\ \\ \\ =\dfrac{\mathrm{sen\,}x}{\cos x} \cdot \cos x\left(1+\dfrac{\cos x}{\mathrm{sen\,}x} \right )\\ \\ \\ =\mathrm{sen\,}x\left(1+\dfrac{\cos x}{\mathrm{sen\,}x} \right )\\ \\ \\ =\mathrm{sen\,}x+\mathrm{sen\,}x\cdot \dfrac{\cos x}{\mathrm{sen\,}x}\\ \\ \\ =\mathrm{sen\,}x+\cos x$