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If tan θ = x − y , the value of sinθ is equal to [If 0° ≤ θ ≤ 90°] x + y
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x − y √2(x2 + y2) -
x + y √2(x2 + y2) -
x + y √2(x2 − y2) -
x − y √2(x2 − y2)
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Correct Option: A
Here,
tan θ = | ||
x + y |
Consider ∆ABC,
Using pythagoras theorem, we get
AC² = AB² + BC²
⇒ AC² = (x + y)² + (x – y)²
= x² + y² + 2xy + x² + y² – 2xy
AC² = 2 (x² + y²)
AC = √2(x² + y²)
As θ lies in first quadrant,
∵ sinq will be +ve
sinθ = | ||
AC |
sinθ = | ||
√2(x² + y²) |