The equation $${\cos ^2}\theta $$ = $$\frac{{{{\left( {x + y} \right)}^2}}}{{4xy}}$$ is only possible when ?
A. x = -y
B. x > y
C. x = y
D. x < y
Answer: Option C
Solution (By Examveda Team)
$$\eqalign{ & {\cos ^2}\theta = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & {\text{Max value of }}{\cos ^2}\theta = 1 \cr & \Rightarrow 1 = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr & \Rightarrow 4xy = {\left( {x + y} \right)^2} \cr & \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr & \Rightarrow 0 = {x^2} + {y^2} - 2xy \cr & \Rightarrow 0 = {\left( {x - y} \right)^2} \cr & \Rightarrow 0 = x - y \cr & \Rightarrow x = y \cr} $$This Question Belongs to Arithmetic Ability >> Trigonometry
Can anyone please tell me why are we considering the max value of cos²θ? why not minimum or the other values?