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JEE Main 2019 · 9 Apr · Shift I

JEE Main 2019 Mathematics question on Trigonometry

The value of cos⁡210°−cos⁡10°cos⁡50°+cos⁡250°\cos^2 10° - \cos 10° \cos 50° + \cos^2 50° is
  1. A.
    32(1+cos⁡20°)\frac{3}{2}(1 + \cos 20°)
  2. B.
    34+cos⁡20°\frac{3}{4} + \cos 20°
  3. C.
    3/23/2
  4. D.
    3/43/4
Show answer and solution

Correct option: D

Solution

Use the identity: cos²A + cos²B - cosA·cosB where A=10°, B=50°.

Write cos²10° = (1 + cos20°)/2, cos²50° = (1 + cos100°)/2, and cos10°·cos50° = (1/2)[cos40° + cos60°] = (1/2)[cos40° + 1/2].

Sum = (1+cos20°)/2 + (1+cos100°)/2 − (1/2)(cos40° + 1/2)

= 1 + (cos20° + cos100°)/2 − cos40°/2 − 1/4

cos20° + cos100° = 2cos60°cos40° = cos40°, so (cos20°+cos100°)/2 = cos40°/2.

= 1 + cos40°/2 − cos40°/2 − 1/4 = 3/4.

[JEE Main 2019 · 9 Apr · Shift I]

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