Examveda The value of $$\frac{{{\text{sin 6}}{5^ \circ }}}{{\cos {{25}^ \circ }}}$$ is? B. 1C. 2D. No definedAnswer: Option B Solution (By Examveda Team) $$\eqalign{ & = \frac{{{\text{sin 6}}{5^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = \frac{{{\text{sin }}\left( {{{90}^ \circ } - {\text{6}}{5^ \circ }} \right)}}{{\cos {{25}^ \circ }}} \cr & \left[ {\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta } \right] \cr & = \frac{{\cos {{25}^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = 1 \cr} $$ This Question Belongs to Arithmetic Ability >> Trigonometry
Solution (By Examveda Team) $$\eqalign{ & = \frac{{{\text{sin 6}}{5^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = \frac{{{\text{sin }}\left( {{{90}^ \circ } - {\text{6}}{5^ \circ }} \right)}}{{\cos {{25}^ \circ }}} \cr & \left[ {\sin \left( {{{90}^ \circ } - \theta } \right) = \cos \theta } \right] \cr & = \frac{{\cos {{25}^ \circ }}}{{\cos {{25}^ \circ }}} \cr & = 1 \cr} $$
The equation $${\cos ^2}\theta $$ = $$\frac{{{{\left( {x + y} \right)}^2}}}{{4xy}}$$ is only possible when ? A. x = -yB. x > yC. x = yD. x < y View Answer
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