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This question belongs to Arithmetic Ability Trigonometry
Trigonometry
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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

Answer & Solution
Correct Answer: Option C
$$\eqalign{ & \left[ {\frac{{{{\sin }^2}{{24}^ \circ } + {{\sin }^2}{{66}^ \circ }}}{{{{\cos }^2}{{24}^ \circ } + {{\cos }^2}{{66}^ \circ }}} + {{\sin }^2}{{61}^ \circ } + \cos {{61}^ \circ }\sin {{29}^ \circ }} \right] \cr & = \frac{1}{1} + {\sin ^2}{61^ \circ } + \cos {61^ \circ }\sin \left( {{{90}^ \circ } - {{61}^ \circ }} \right) \cr & = 1 + {\sin ^2}{61^ \circ } + {\cos ^2}{61^ \circ } \cr & = 1 + 1 \cr & = 2 \cr} $$
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