If $an heta=frac{x-y}{x+y}$, the value of $sin heta$ is equal to [If $0^{circ} leq heta leq 90^{circ}$ ]

Here, $an heta=frac{x-y}{x+y}$ Consider $Delta mathrm{ABC}$,

Using pythagoras theorem, we get

$mathrm{AC}^{2}=mathrm{AB}^{2}+mathrm{BC}^{2}$

$Rightarrow mathrm{AC}^{2}=(x+y)^{2}+(x-y)^{2}$

$=x^{2}+y^{2}+2 x y+x^{2}+y^{2}-2 x y$

$mathrm{AC}^{2}=2left(x^{2}+y^{2} ight)$

$mathrm{AC}=sqrt{2left(x^{2}+y^{2} ight)}$

As θ lies in first quadrant,

sinθ will be +ve

$sin heta=frac{mathrm{BC}}{mathrm{AC}}$

$sin heta=frac{x-y}{sqrt{2left(x^{2}+y^{2} ight)}}$