Question

Simplify: ${\left(\frac{\sin {47}^{\circ }}{\cos {43}^{\circ }}\right)}^2+{\left(\frac{\cos {43}^{\circ }}{\sin {47}^{\circ }}\right)}^2-4{\cos }^2{45}^{\circ }$

Answer 1

${\left(\frac{\sin {47}^{\circ }}{\cos {43}^{\circ }}\right)}^2+{\left(\frac{\cos {43}^{\circ }}{\sin {47}^{\circ }}\right)}^2-4{\cos }^2{45}^{\circ }$

$={\left(\frac{\sin (90-43{)}^{\circ }}{\cos {43}^{\circ }}\right)}^2+{\left(\frac{\cos (90-47{)}^{\circ }}{\sin {47}^{\circ }}\right)}^2-4{\cos }^2{45}^{\circ }$

$={\left(\frac{\cos {43}^{\circ }}{\cos {43}^{\circ }}\right)}^2+{\left(\frac{\sin {47}^{\circ }}{\sin {47}^{\circ }}\right)}^2-4{\cos }^2{45}^{\circ }$

[$\because \sin ({90}^{\circ }-\theta )=\cos \theta ,\cos ({90}^{\circ }-\theta )=\sin \theta$]

$=1+1-4(\frac{1}{\sqrt{2}}{)}^2$

$=2-4\left(\frac{1}{2}\right)$

$=2-2$

$=0$

$\therefore {\left(\frac{\sin {47}^{\circ }}{\cos {43}^{\circ }}\right)}^2+{\left(\frac{\cos {43}^{\circ }}{\sin {47}^{\circ }}\right)}^2-4{\cos }^2{45}^{\circ }=0$