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The value of $$\frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ } + 4{\text{se}}{{\text{c}}^2}{{30}^ \circ } - {\text{ta}}{{\text{n}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {\text{co}}{{\text{s}}^2}{{30}^ \circ }}}$$      is?

A. $$\frac{{64}}{{\sqrt 3 }}$$

B. $$\frac{{55}}{{12}}$$

C. $$\frac{{67}}{{12}}$$

D. $$\frac{{67}}{{10}}$$

Answer: Option B

Solution(By Examveda Team)

$$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ } + 4{\text{se}}{{\text{c}}^2}{{30}^ \circ } - {\text{ta}}{{\text{n}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {\text{co}}{{\text{s}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{2}} \right)}^2} + 4{{\left( {\frac{2}{{\sqrt 3 }}} \right)}^2} - 1}}{1} \cr & \left( {{{\sin }^2}{\text{A}} + {{\cos }^2}{\text{A}} = {\text{1}}} \right) \cr & \Rightarrow \frac{1}{4} + \frac{{4 \times 4}}{3} - 1 \cr & \Rightarrow \frac{1}{4} + \frac{{16}}{3} - 1 \cr & \Rightarrow \frac{{3 + 64 - 12}}{{12}} \cr & \Rightarrow \frac{{55}}{{12}} \cr} $$

This Question Belongs to Arithmetic Ability >> Trigonometry

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