Question

${\left(\frac{\sin {50}^{\circ }}{\cos {40}^{\circ }}\right)}^2+{\left(\frac{\cos {28}^{\circ }}{\sin {62}^{\circ }}\right)}^2-2{\tan }^2{45}^{\circ }$

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $-2$

Answer 1

Answer: (A)

${\left(\frac{\sin {50}^{\circ }}{\cos {40}^{\circ }}\right)}^2+{\left(\frac{\cos {28}^{\circ }}{\sin {62}^{\circ }}\right)}^2-2{\tan }^2{45}^{\circ }$

= ${\left(\frac{\sin (90-40{)}^{\circ }}{\cos {40}^{\circ }}\right)}^2+{\left(\frac{\cos {28}^{\circ }}{\sin (90-28{)}^{\circ }}\right)}^2-2{\tan }^2{45}^{\circ }$

= ${\left(\frac{\cos {40}^{\circ }}{\cos {40}^{\circ }}\right)}^2+{\left(\frac{\cos {28}^{\circ }}{\cos {28}^{\circ }}\right)}^2-2{\tan }^2{45}^{\circ }$

= $1+1-2(1{)}^2$

= $0$