If tan theta-cot theta=a and cos theta+sin theta=b, then show that (a^2+4)(b^2-1)^2=4.
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Solution:
We have x2 + y2 = (a sin θ + b cos θ)2 + (a cos θ – b sin θ)2
= (a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ) + (a2 cos2 θ + b2 sin2 θ - 2ab sin θ cos θ)
= a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ + a2 cos2 θ + b2 sin2 θ - 2ab sin θ cos θ
= a2 sin2 θ + b2 cos2 θ + a2 cos2 θ + b2 sin2 θ
= a2 sin2 θ + a2 cos2 θ + b2 sin2 θ + b2 cos2 θ
= a2 (sin2 θ + cos2 θ) + b2 (sin2 θ + cos2 θ)
= a2 (1) + b2 (1); [since, sin2 θ + cos2 θ = 1]
= a2 + b2
Therefore, x2 + y2 = a2 + b2
which is the required θ-eliminate.
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