Examveda What is the value of (1 + cot2θ)(1 - cos2θ). A. 1B. Can not definedD. $$\frac{1}{2}$$Answer: Option A Solution (By Examveda Team) $$\eqalign{ & \left( {1 + {{\cot }^2}\theta } \right)\left( {1 - {{\cos }^2}\theta } \right) \cr & = \left( {1 + \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right){\sin ^2}\theta \cr & = \left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right){\sin ^2}\theta \cr & = 1 \cr} $$ This Question Belongs to Arithmetic Ability >> Trigonometry
Solution (By Examveda Team) $$\eqalign{ & \left( {1 + {{\cot }^2}\theta } \right)\left( {1 - {{\cos }^2}\theta } \right) \cr & = \left( {1 + \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right){\sin ^2}\theta \cr & = \left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right){\sin ^2}\theta \cr & = 1 \cr} $$
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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