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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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.