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JEE Main 2013

JEE Main 2013 Mathematics question on Trigonometry

ABCDABCD is a trapezium such that ABAB and CDCD are parallel and BC⊥CDBC \perp CD, if ∠ADB=θ\angle ADB = \theta, BC=pBC = p and CD=qCD = q, then ABAB is equal to
  1. A.
    (p2+q2)sin⁡θpcos⁡θ+qsin⁡θ\dfrac{(p^2 + q^2)\sin\theta}{p\cos\theta + q\sin\theta}
  2. B.
    p2+q2cos⁡θpcos⁡θ+qsin⁡θ\dfrac{p^2 + q^2\cos\theta}{p\cos\theta + q\sin\theta}
  3. C.
    p2+q2p2cos⁡θ+q2sin⁡θ\dfrac{p^2 + q^2}{p^2\cos\theta + q^2\sin\theta}
  4. D.
    (p2+q2)sin⁡θ(pcos⁡θ+qsin⁡θ)2\dfrac{(p^2 + q^2)\sin\theta}{(p\cos\theta + q\sin\theta)^2}
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Correct option: A

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