Math, asked by AnkurGav, 10 months ago

If tan theta-cot theta=a and cos theta+sin theta=b, then show that (a^2+4)(b^2-1)^2=4.​

Answers

Answered by kaynat87
2

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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