Question

Si senx.secx=√5.cosx.cscx, calcule la expresión



M=103 ( 3sen^8 x + 2cos^8 x / 4sen^8 x + 3cos^8 x )

Answer 1

$\sin x\cdot \sec x =\sqrt{5}\cdot \cos x \cdot \csc x \\[4pt] \sin x\cdot \dfrac{1}{\cos x} =\sqrt{5}\cdot \cos x \cdot \dfrac{1}{\sin x} \ \Rightarrow \ \sin^2x =\sqrt{5}\cos^2x \\[4pt] \text{Elevando a la cuarta potencia} \\[4pt] (\sin^2x)^4 =(\sqrt{5}\cos^2x)^4 \\[4pt] \boxed{\sin^8x =25\cos^8x} \ \ \leftarrow \ \text{Esto se va reemplazar en }M$

$M=103\left[ \dfrac{3\sin^8x+2\cos^8x}{4\sin^8x+3\cos^8x} \right] \\[6pt] M=103\left[\dfrac{3(25\cos^8x)+2\cos^8x}{4(25\cos^8x)+3\cos^8x}\right]=103\left[\dfrac{75\cos^8x+2\cos^8x}{100\cos^8x+3\cos^8x}\right]\\[6pt] M=103\left[\dfrac{77\cos^8x}{103\cos^8x}\right]=77 \ \ \leftarrow \ Respuesta.$