The value of $$\left[ {\frac{{{{\sin }^2}{{24}^ \circ } + {{\sin }^2}{{66}^ \circ }}}{{{{\cos }^2}{{24}^ \circ } + {{\cos }^2}{{66}^ \circ }}} + {{\sin }^2}{{61}^ \circ } + \cos {{61}^ \circ }\sin {{29}^ \circ }} \right]$$ is equal to
A. 1
B. 3
C. 2
D. 0
Answer: Option C
Solution (By Examveda Team)
$$\eqalign{ & \left[ {\frac{{{{\sin }^2}{{24}^ \circ } + {{\sin }^2}{{66}^ \circ }}}{{{{\cos }^2}{{24}^ \circ } + {{\cos }^2}{{66}^ \circ }}} + {{\sin }^2}{{61}^ \circ } + \cos {{61}^ \circ }\sin {{29}^ \circ }} \right] \cr & = \frac{1}{1} + {\sin ^2}{61^ \circ } + \cos {61^ \circ }\sin \left( {{{90}^ \circ } - {{61}^ \circ }} \right) \cr & = 1 + {\sin ^2}{61^ \circ } + {\cos ^2}{61^ \circ } \cr & = 1 + 1 \cr & = 2 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y

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