The value of $$\frac{{\sin \left( {{{78}^ \circ } + \theta } \right) - \cos \left( {{{12}^ \circ } - \theta } \right) + \left( {{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}{{20}^ \circ }} \right)}}{{\sin {{25}^ \circ }\cos {{65}^ \circ } + \cos {{25}^ \circ }\sin {{65}^ \circ }}}{\text{ is:}}$$
A. 2
B. -1
C. -2
Answer: Option B
Solution (By Examveda Team)
$$\eqalign{ & \frac{{\sin \left( {{{78}^ \circ } + \theta } \right) - \cos \left( {{{12}^ \circ } - \theta } \right) + \left( {{{\tan }^2}{{70}^ \circ } - {\text{cose}}{{\text{c}}^2}{{20}^ \circ }} \right)}}{{\sin {{25}^ \circ }\cos {{65}^ \circ } + \cos {{25}^ \circ }\sin {{65}^ \circ }}} \cr & \frac{{\sin \left( {{{78}^ \circ } + \theta } \right) - \sin \left( {{{78}^ \circ } + \theta } \right) + \left( {{{\tan }^2}{{70}^ \circ } - {{\sec }^2}{{70}^ \circ }} \right)}}{{\sin {{25}^ \circ }\sin {{25}^ \circ } + \cos {{25}^ \circ }\sin {{25}^ \circ }}} \cr & \frac{{ - 1}}{{{{\sin }^2}{{25}^ \circ } + {{\cos }^2}{{25}^ \circ }}} \cr & = - 1 \cr} $$This Question Belongs to Arithmetic Ability >> Trigonometry
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