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JEE Main 20199 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
9 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2019 · 9 April · Morning Shift · Mathematics · Trigonometric Ratios & Identities

The value of ^ 2 10 ° – 10 ° 50 ° + co s ^ 2 50 ° is

The value of cos⁡210°–cos⁡10°cos⁡50°+cos250°\cos^{2}10 ° – \cos 10 ° \cos 50 ° + co{s}^{2}50 ° is

Options

  1. A34\dfrac{3}{4}
  2. B34+cos⁡20°\dfrac{3}{4} + \cos 20 °
  3. C32\dfrac{3}{2}
  4. D32(1+cos⁡20°)\dfrac{3}{2} ( 1 + \cos 20 ° )

Correct answer

A. 34\dfrac{3}{4}

Step-by-step solution

cos⁡210°+cos⁡250°−cos⁡10°cos⁡50°\cos^{2}10 ° + \cos^{2}50 ° - \cos 10 ° \cos 50 °

Using, cos⁡2x=2cos⁡2x−1,\cos 2x = 2\cos^{2}x - 1, ⇒cos⁡2x=1+cos⁡2x2,⇒ \cos^{2}x = \dfrac{1 + \cos 2x}{2}, we get

=12[1+cos⁡20°+1+cos⁡100°−2cos⁡50°cos⁡10°]= \dfrac{1}{2}\left[1 + \cos 20 ° + 1 + \cos 100 ° - 2\cos 50 ° \cos 10 °\right]

=12[2+cos⁡100°+cos⁡20°−cos⁡60°−cos⁡40°]= \dfrac{1}{2}\left[2 + \cos 100 ° + \cos 20 ° - \cos 60 ° - \cos 40 °\right]

=12[2+cos⁡100°+cos⁡20°−12−cos⁡40°]= \dfrac{1}{2}\left[2 + \cos 100 ° + \cos 20 ° - \dfrac{1}{2} - \cos 40 °\right]

Using, cosC+cosD=2cos⁡(C+D2)cos⁡(C−D2),cosC + cosD = 2\cos \left(\dfrac{C + D}{2}\right)\cos \left(\dfrac{C - D}{2}\right), we get

=12[32+2cos60°cos⁡40°−cos⁡40°]= \dfrac{1}{2}\left[\dfrac{3}{2} + 2cos {60}^{°}\cos 40 ° - \cos 40 °\right]

=12[32+2×12×cos⁡40°−cos⁡40°]= \dfrac{1}{2}\left[\dfrac{3}{2} + 2 \times \dfrac{1}{2} \times \cos 40 ° - \cos 40 °\right]

=12[32+cos⁡40°−cos⁡40°]= \dfrac{1}{2}\left[\dfrac{3}{2} + \cos 40 ° - \cos 40 °\right]

=12×32=34.= \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4} .

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