Question

$(\cos 0^{\circ }+\sin {45}^{\circ }+\sin {30}^{\circ })(\sin {90}^{\circ }-\cos {45}^{\circ }+\cos {60}^{\circ })=$

• A. $\frac{5}{4}$
• B. $\frac{9}{4}$
• C. $\frac{7}{4}$
• D. None of these

Answer 1

Answer: (C)

$\frac{7}{4}$

$\left(\cos 0^o+\sin {45}^o+\sin {30}^o\right)\left(\sin {90}^o-\cos {45}^o+\cos {60}^o\right)$

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

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

$=(\frac{3}{2}+\frac{1}{\sqrt{2}})(\frac{3}{2}-\frac{1}{\sqrt{2}})$

$={\left(\frac{3}{2}\right)}^2-{\left(\frac{1}{\sqrt{2}}\right)}^2$

$=\frac{9}{4}-\frac{1}{2}=\frac{7}{4}$