The numerical value of $$\frac{{{\text{co}}{{\text{s}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{60}^ \circ }}}$$ + $$\frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ }}}{{{{\sin }^2}{{45}^ \circ }}}$$ - $$\frac{{{\text{ta}}{{\text{n}}^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{45}^ \circ }}}$$ - $$\frac{{{{\sin }^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{30}^ \circ }}}$$ is?
A. $$\frac{3}{4}$$
B. $$\frac{1}{4}$$
C. $$\frac{1}{2}$$
D. $${\text{1}}\frac{1}{4}$$
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{\text{co}}{{\text{s}}^2}{{45}^ \circ }}}{{{{\sin }^2}{{60}^ \circ }}}{\text{ + }}\frac{{{\text{co}}{{\text{s}}^2}{{60}^ \circ }}}{{{{\sin }^2}{{45}^ \circ }}} - \frac{{{\text{ta}}{{\text{n}}^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{45}^ \circ }}} - \frac{{{{\sin }^2}{{30}^ \circ }}}{{{\text{co}}{{\text{t}}^2}{{30}^ \circ }}} \cr & \Rightarrow \frac{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{{{\left( {\frac{{\sqrt 3 }}{2}} \right)}^2}}} + \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}} - \frac{{{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( 1 \right)}^2}}} - \frac{{{{\left( {\frac{1}{2}} \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2}}} \cr & \Rightarrow \left( {\frac{1}{2} \times \frac{4}{3}} \right) + \left( {\frac{1}{4} \times \frac{2}{1}} \right) - \left( {\frac{1}{3} \times 1} \right) - \left( {\frac{1}{4} \times \frac{1}{3}} \right) \cr & \Rightarrow \frac{2}{3} + \frac{1}{2} - \frac{1}{3} - \frac{1}{{12}} \cr & \Rightarrow \frac{1}{3} + \frac{1}{2} - \frac{1}{{12}} \cr & \Rightarrow \frac{{4 + 6 - 1}}{{12}} \cr & \Rightarrow \frac{9}{{12}} \cr & \Rightarrow \frac{3}{4} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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