1y(x2 + y2)
xy
1y√x2+y2
x−yy
Correct answer is C
cotθ=xy
⟹tanθ=yx
opp=y;adj=x
Using Pythagoras theorem, Hyp2=Opp2+Adj2
Hyp2=y2+x2
Hyp=√y2+x2
sinθ=y√y2+x2
∴
= \frac{1}{y}(\sqrt{y^{2} + x^{2}})
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