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°

= cot

^{2}θ – cosec^{2}θ +√3 tan 60° tan 12° tan(90-12)°= – (cosec

^{2}θ – cot^{2}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.