Question

necesito ayuda porfavor

Answer 1

Respuesta:

CALCULAR:

$\boxed{\mathrm{E=\frac{\sin \left75^{\circ \:}\right\cos \left30^{\circ \:}\right-\sin \left30^{\circ \:}\right\cdot\cos \left75^{\circ \:}\right}{\cos \left60^{\circ \:}\right\cdot\cos \left15^{\circ \:}\right+\sin \left15^{\circ \:}\right\cdot\sin \left60^{\circ \:}\right}}}$

Resolvemos:

$\sin \left75^{\circ \:}\right\cos \left30^{\circ \:}\right-\sin \left30^{\circ \:}\right\cdot\cos \left75^{\circ \:}\right=\sin 75^{\circ \:}-30^{\circ \:}$

$\sin45^{\circ \:}=\frac{\sqrt{2}}{2}$

$\cos \left60^{\circ \:}\right\cdot\cos \left15^{\circ \:}\right+\sin \left15^{\circ \:}\right\cdot\sin \left60^{\circ \:}\right=\cos 60^{\circ \:}-15^{\circ \:}$

$\cos 45^{\circ \:}=\frac{\sqrt{2}}{2}$

$\sin \left75^{\circ \:}\right\cos \left30^{\circ \:}\right-\sin \left30^{\circ \:}\right\cdot\cos \left75^{\circ \:}\right=\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2+\sqrt{3}}}{2}-\frac{1}{2}\cdot \frac{\sqrt{2-\sqrt{3}}}{2}$

$\cos \left60^{\circ \:}\right\cdot\cos \left15^{\circ \:}\right+\sin \left15^{\circ \:}\right\cdot\sin \left60^{\circ \:}\right=\frac{1}{2}\cdot \frac{\sqrt{2+\sqrt{3}}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2-\sqrt{3}}}{2}$

$\boxed{\mathrm{E=\frac{\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2+\sqrt{3}}}{2}-\frac{1}{2}\cdot \frac{\sqrt{2-\sqrt{3}}}{2}}{\frac{1}{2}\cdot \frac{\sqrt{2+\sqrt{3}}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2-\sqrt{3}}}{2}}}}$

Simplificamos:

$\mathrm{Multiplicar \ por \ el \ conjugado:}$

$\frac{\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}}$

$\frac{\left(\sqrt{3}\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}\right)}{\left(\sqrt{2+\sqrt{3}}+\sqrt{3}\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}\right)}$

$\left(\sqrt{3}\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}\right)=4\sqrt{3}-4$

$\left(\sqrt{2+\sqrt{3}}+\sqrt{3}\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}\right)=4\sqrt{3}-4$

$\frac{4\sqrt{3}-4}{4\sqrt{3}-4}=1$

Solución:  Respuesta correcta (Opción A)

$\boxed{\mathrm{E=1}}$

Saludos...