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Rajan@2021
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Asked: 2021-04-08T16:56:22+05:30 In:

Prove that: [sec(90°-θ) cosecθ – tan(90°-θ) cotθ + cos²25° + cos²65°]/3tan27°tan63° = 2/3

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This question is from trigonometry topic – trigonometric ratios on complementary angles in which we have been asked to prove that [sec(90°-θ) cosecθ – tan(90°-θ) cotθ + cos²25° + cos²65°]/3tan27°tan63° = 2/3

RS Aggarwal, Class 10, Chapter 12, question no 5(vi)

cbse class 10rs aggarwaltrigonometric ratios
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    1. Guru
      Prove that: [sec(90°-θ) cosecθ – tan(90°-θ) cotθ + cos²25° + cos²65°]/3tan27°tan63° = 2/3

      Hope it helps you.

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