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The equation $${\cos ^2}\theta $$ = $$\frac{{{{\left( {x + y} \right)}^2}}}{{4xy}}$$ is only possible when ?
Answer & Solution
Correct Answer:
Option
C
$$\eqalign{
& {\cos ^2}\theta = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr
& {\text{Max value of }}{\cos ^2}\theta = 1 \cr
& \Rightarrow 1 = \frac{{{{\left( {x + y} \right)}^2}}}{{4xy}} \cr
& \Rightarrow 4xy = {\left( {x + y} \right)^2} \cr
& \Rightarrow 4xy = {x^2} + {y^2} + 2xy \cr
& \Rightarrow 0 = {x^2} + {y^2} - 2xy \cr
& \Rightarrow 0 = {\left( {x - y} \right)^2} \cr
& \Rightarrow 0 = x - y \cr
& \Rightarrow x = y \cr} $$
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