Question

Sec2theta =4xy / (x+y)2 Is true if and only if ______

Answer 1

Answer:

if x=y, and x≠0

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Answer 2

Answer →

${sec}^{2} \theta = \frac{4xy}{ {(x + y)}^{2} }$

is true if only if →

x ≠ 0 , y ≠ 0 , x + y = 0

Step by step Explanation →

${sec}^{2} \theta = \frac{4xy}{ {(x + y)}^{2} }$

We know that →

• ${sec}^{2} \theta \geqslant 1$

So,

$\frac{4xy}{ {(x + y)}^{2} } \geqslant 1$

We can write it as →

${(x + y)}^{2} \leqslant 4xy$

${(x + y)}^{2} - 4xy \leqslant 0$

${(x - y)}^{2} \leqslant 0$

For the real value of x and y →

${(x - y)}^{2} \geqslant 0$

Which means that →

${(x - y)}^{2} = 0$

So we can say that →

$x = y$

x ≠ 0 , y ≠ 0

__________________________

Hence Proved !!

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