Question

if cosec^2 theta = 4xy/(x+y)^2, then find the relation between x and y

Answer 1

cosec2Ѳ = 4xy / (x+y)2

Let cosec2Ѳ has a real value

Therefore, cosec2Ѳ ≥ 1

So, 4xy/(x+y)2 ≥ 1

or (x+y)2 ≤ 4xy

or (x+y)2 - 4xy ≤ 0

or (x-y)2 ≤ 0

This implies that x=y

Therefore given equation holds only if x=y

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Answer 2

The relation between x and y is x = y   if cosec²θ = 4xy/(x + y)²

Given:

• cosec²θ = 4xy/(x + y)²

To Find:

• The relation between x and y

Solution:

• cosec²θ = 1 + cot²θ
• square of a real number is non negative
• (x + y)² = x² + y²  + 2xy
• (x - y)² = x² + y² - 2xy

cosec²θ = 4xy/(x + y)²

Step 1:

Substract 1 from both sides

cosec²θ - 1 = 4xy/(x + y)² - 1

Step 2:

Use cosec²θ - 1  = cot²θ and simplify RHS

cot²θ = (4xy - (x² + y² +2xy))/(x + y)²  

cot²θ = -( x² + y² - 2xy))/(x + y)²  

cot²θ = -( x-y)²/(x + y)²  

cot²θ = - ((x-y)/(x + y))²  

cot²θ +  ((x-y)/(x + y))² = 0

Step 3:

As sum of 2 square terms is zero hence both terms should be zero

=>  ((x-y)/(x + y))² = 0

=> x - y = 0      

=> x = y

The relation between x and y is x = y