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