This question is from trigonometry topic -trigonometric ratios on complementary angles in which we have been asked to prove that cosec(65°+θ) – sec(25°-θ) – tan(55-θ) + cot(35°+θ) = 0
RS Aggarwal, Class 10, chapter 12, question no 7(iv)
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LHS=cosec(65°+θ)−sec[90°−(65°+θ)]−tan[90°−(35°+θ)]+cot(35°+θ)
=cosec(65°+θ)−cosec(65°+θ)−cot(35°+θ)+cot(35°+θ)
=0
=RHS