Question

The distance between the point (acos theta + bcos theta , 0) and (0, asin theta - bcos theta) is
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Answer 1

Answer:

Given the point A(cosθ+bsinθ,0),(0,asinθ−bcosθ)

By distance formula

The distance of AB

under root (x2−x1)2+(y2−y1)2

= under root[0−(acosθ+bsinθ)2+(asinθ−bcosθ−0]2= under root a2cos2θ+2abcosθsinθ+a2sin2θ+b2cos2θ−2absinθcosθ

= under root (a2+b2)cos2θ+(a2+b2)sin2θ

= under root a2+b2

[∵cos2θ+sin2θ=1]

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

The points are (a cos ∅ + b sin ∅, 0) and (0, a sin ∅ - b cos∅)

By distance formula,

√(a cos∅ + b sin ∅ - 0)² + (0-(a sin ∅ - b cos ∅))²

using identity (a±b)² = a² + b² ±2ab,

√ a²cos² ∅ + b²sin² ∅ + 2ab sin∅ cos∅ + b²cos²∅ + a²sin²∅ - 2ab sin∅ cos∅

√ a²cos²∅ + a²sin²∅ + b²cos²∅ + b²sin²∅

√ a²( cos²∅+ sin²∅ ) + b²(cos²∅ + sin²∅)

using identity sin² ∅ + cos²∅ = 1,

√ a² (1) + b² (1)

=> √a²+b²

hope it helps………)