This question is from trigonometry topic -trigonometric ratios on complementary angles in which we have been asked to prove that sin(70°+θ) – cos(20°-θ) = 0
RS Aggarwal, Class 10, chapter 12, question no 7(i)
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LHS=sin(70°+θ)−cos(20°+θ)
=sin(70°+θ)−cos[90°−(70°+θ)]
=sin(70°+θ)−sin(70°+θ)
=0=RHS