Question

(PUC-SP) sendo seno x =1/3 ,com 0 < x< pi/2, calcule (senx.cosx-tgx)/1-cossecx

Answer 1

 Encontremos o cosseno de x fazendo uso da equação $\sin^2 x + \cos^2 x = 1$. Segue,

$\\ \sin^2 x + \cos^2 x = 1 \\\\ \left ( \frac{1}{3} \right )^2 + \cos^2 x = 1 \\\\ \cos^2 x = 1 - \frac{1}{9} \\\\ \cos^2 x = \frac{8}{9} \\\\ \cos x = \pm \frac{2\sqrt{2}}{3}$

 Mas, x está no primeiro quadrante, então tiramos que $\boxed{\cos x = + \frac{2\sqrt{2}}{3}}$.

 Por conseguinte, substituímos...

$\\ \frac{\sin x \cdot \cos x - \tan x}{1 - \csc x} = \\\\\\ \frac{\sin x \cdot \cos x - \frac{\sin x}{\cos x}}{1 - \frac{1}{\sin x}} = \\\\\\ \left ( \frac{\sin x \cdot \cos^2 x - \sin x}{\cos x} \right ) \div \left ( \frac{\sin x - 1}{\sin x} \right ) = \\\\\\ \frac{\sin x \cdot \cos^2 x - \sin x}{\cos x} \times \frac{\sin x}{\sin x - 1} = \\\\\\ \frac{\frac{1}{3} \cdot \frac{8}{9} - \frac{1}{3}}{\frac{2\sqrt{2}}{3}} \times \frac{\frac{1}{3}}{\frac{1}{3} - 1} = \\\\\\ \frac{- \frac{1}{27}}{\frac{2\sqrt{2}}{3}} \times \frac{\frac{1}{3}}{- \frac{2}{3}} = \\\\\\ - \frac{1}{27} \times \frac{3}{2\sqrt{2}} \times \frac{1}{3} \times - \frac{3}{2} = \\\\\\ \frac{1}{36\sqrt{2}} = \\\\\\ \boxed{\frac{\sqrt{2}}{72}}$