Question

If $A={45}^{\circ }and{30}^{\circ },$ then the value of $\frac{2si{n}^2({90}^{\circ }-B)+sec({90}^{\circ }-A)}{cosec({90}^{\circ }-B)+ta{n}^2(3B-A)}$ is

• A.
  $\frac{\sqrt{3}(2\sqrt{2}+3)}{2(2+\sqrt{3})}$
• B.
  $\frac{2\sqrt{2}-3}{2-\sqrt{3}}$
• C.
  $\frac{2+\sqrt{3}}{\sqrt{3}(2\sqrt{2}+3)}$
• D.
  1

Answer 1

The correct option is A

$\frac{\sqrt{3}(2\sqrt{2}+3)}{2(2+\sqrt{3})}$
Given $A={45}^{\circ }andB={30}^{\circ }\therefore \frac{2si{n}^2({90}^{\circ }-B)sec({90}^{\circ }-A)}{cosec({90}^{\circ }-B)+ta{n}^2(3B-A)}=\frac{2si{n}^2{60}^{\circ }+sec{45}^{\circ }}{cosec{60}^{\circ }+ta{n}^2{45}^{\circ }}=\frac{2\times {\left(\frac{\sqrt{3}}{2}\right)}^2+\sqrt{2}}{\frac{2}{\sqrt{3}}+(1{)}^2}=\frac{3+2\sqrt{2}}{2(\frac{2}{\sqrt{3}}+1)}=\frac{3+2\sqrt{2}}{4+2\sqrt{3}}\times \frac{\sqrt{3}}{1}=\frac{\sqrt{3}(3+2\sqrt{2})}{2(2+\sqrt{3})}$

Hence, the correct answer is option a.