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JEE Main 2016 · Mathematics · Trigonometry and Inverse Trigonometric Functions · hard

The value of (√3)/(sin 20^°)-(1)/(cos 20^°) is

Single correct option

The value of 3sin⁡20∘−1cos⁡20∘\dfrac{\sqrt3}{\sin20^\circ}-\dfrac{1}{\cos20^\circ} is
  1. A. 22
  2. B. 44
  3. C. 232\sqrt3
  4. D. 88
Show answer and solution

Answer: (B) 44

Combine the fractions: 3cos⁡20∘−sin⁡20∘sin⁡20∘cos⁡20∘\dfrac{\sqrt3\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}. Write the numerator as 2(32cos⁡20∘−12sin⁡20∘)=2sin⁡(60∘−20∘)=2sin⁡40∘2\left(\tfrac{\sqrt3}{2}\cos20^\circ-\tfrac12\sin20^\circ\right)=2\sin(60^\circ-20^\circ)=2\sin40^\circ. For the denominator use sin⁡20∘cos⁡20∘=12sin⁡40∘\sin20^\circ\cos20^\circ=\tfrac12\sin40^\circ. So the expression equals 2sin⁡40∘12sin⁡40∘=4\dfrac{2\sin40^\circ}{\tfrac12\sin40^\circ}=4. A numerical check gives 4.00004.0000. The auxiliary-angle step (writing 3cos⁡θ−sin⁡θ\sqrt3\cos\theta-\sin\theta as 2sin⁡(60∘−θ)2\sin(60^\circ-\theta)) is the key idea.
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