New Enroll in RBI Grade B New Batch - Starts on May 21

    Question

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    A sinθ.sec²θ Correct Answer Incorrect Answer
    B cos ² θ.cot² θ Correct Answer Incorrect Answer
    C secθ/cosθ Correct Answer Incorrect Answer
    D sinθ.cosθ Correct Answer Incorrect Answer

    Solution

    √(sec2θ + cosec2θ) × √(tan2θ – sin2θ) ⇒ √(1/cos2θ + 1/sin2θ) × √(sin2θ/cos2θ – sin2θ) ⇒ √[(sin2θ + cos2θ) / (sin2θ.cos2θ)] × √[sin2θ.(1/cos2θ – 1) ⇒ √[1/ (sin2θ.cos2θ)] × √[sin2θ.(1 – cos2θ) /cos2θ) ⇒ 1/ (sinθ.cosθ) × sin2θ/cosθ  => sinθ.sec2θ

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