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)
$$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ } + 4{\text{se}}{{\text{c}}^2}{{30}^ \circ } - {\text{ta}}{{\text{n}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{30}^ \circ } + {\text{co}}{{\text{s}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{2}} \right)}^2} + 4{{\left( {\frac{2}{{\sqrt 3 }}} \right)}^2} - 1}}{1} \cr & \left( {{{\sin }^2}{\text{A}} + {{\cos }^2}{\text{A}} = {\text{1}}} \right) \cr & \Rightarrow \frac{1}{4} + \frac{{4 \times 4}}{3} - 1 \cr & \Rightarrow \frac{1}{4} + \frac{{16}}{3} - 1 \cr & \Rightarrow \frac{{3 + 64 - 12}}{{12}} \cr & \Rightarrow \frac{{55}}{{12}} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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