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Rajan@2021
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Prove that: cotθ tan(90°-θ) – sec(90°-θ) cosecθ + √3tan12° tan60° tan78° = 2

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This question is from trigonometry topic -trigonometric ratios on complementary angles in which we have been asked to prove that cotθ tan(90°-θ) – sec(90°-θ) cosecθ + √3tan12° tan60° tan78° = 2

RS Aggarwal, Class 10, chapter 12, question no 5(vii)

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  1. Consider L.H.S.

    = cot θ tan (90° -θ) – sec (90° – θ)cosec θ +√3tan 12°tan 60°tan 78°

    = cot θ cot θ – cosec θ cosec θ +√3 tan 60° tan 12° tan 78°

    = cot2 θ – cosec2 θ +√3 tan 60° tan 12° tan(90-12)°

    = – (cosec2 θ – cot2 θ) +√3 tan 60° tan 12° cot 12°

    = – 1 + √3(√3 × tan 12° cot 12°)

    = – 1 + √3(√3 × 1)

    = – 1 + 3

    = 2 = R.H.S.

    Hence, proved.

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