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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

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Comments ( 1 )

  1. Sirajum Munira
    Sirajum Munira :
    2 years ago

    Can anyone please tell me why are we considering the max value of cos²θ? why not minimum or the other values?

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