Question

The value of
$\frac{co{t}^2{30}^{\circ }+8si{n}^2{45}^{\circ }+\frac{3}{2}se{c}^2{30}^{\circ }+2co{s}^2{90}^{\circ }}{2sec{60}^{\circ }+3cosec{30}^{\circ }-\frac{7}{3}ta{n}^2{60}^{\circ }}$ is :

• A. 4
• B. 2
• C. 3
• D. 1

Answer 1

The correct option is C

3

Given : $\frac{co{t}^2{30}^{\circ }+8si{n}^2{45}^{\circ }+\frac{3}{2}se{c}^2{30}^{\circ }+2co{s}^2{90}^{\circ }}{2sec{60}^{\circ }+3cosec{30}^{\circ }-\frac{7}{3}ta{n}^2{60}^{\circ }}$

Now : ${\cot }^2{30}^{\circ }$ = 3 , ${\sin }^2{45}^{\circ }$ = $\frac{1}{2}$ , ${\sec }^2{30}^{\circ }$ = $\frac{4}{3}$ , ${\cos }^2{90}^{\circ }$ = 0 , $\sec {60}^{\circ }$ = 2 , $cosec{30}^{\circ }$ = 2 , ${\tan }^2{60}^{\circ }$ = 3

Substituting these values in the given equation we get :

$\frac{3+\frac{8}{2}+\frac{3}{2}\times \frac{4}{3}+2\times 0}{2\times 2+3\times 2-\frac{7}{3}\times 3}$

= $\frac{3+4+2+0}{4+6-7}$

= $\frac{9}{3}=3$