If E = ^ 2 71 ^ o + ^ 2 49 ^ o + cos 71 ^ o cos 49 ^ o , then the value of 10E is equal to

Correct answer: A. 7 . 5 E = 1 + cos 142 ^ o 2 + 1 + cos 98 ^ o 2 + (1)/(2) (2cos 71 ^ o cos 49 ^ o ) = 1 + (1)/(2) (cos 142 ^ o + cos 98 ^ o ) + (1)/(2) ( cos 120 ^ o + cos 22 ^ o ) = 1 + (1)/(2) (2cos 120 ^ o cos 22 ^ o ) + (1)/(2)(- (1)/(2) +cos 22 ^ o ) = 1 + (- (1)/(2)) cos 22 ^ o - (1)/(4) + (1)/(2) cos 22 ^ o = (3)/(4) E = (3)/(4) = 0.75 ⇒ 10E = 7.5

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Trigonometric Ratios & Identities

JEE Main · Mathematics · Trigonometric Ratios & Identities

If E = ^ 2 71 ^ o + ^ 2 49 ^ o + cos 71 ^ o cos 49 ^ o , then the...

E = \cos^{2} {71}^{o} + \cos^{2} {49}^{o} + cos {71}^{o}cos {49}^{o},

then the value of

10E

is equal to

Options

A $7 . 5$
B $2 . 5$
C $12 . 5$
D $10 . 5$

Correct answer

A. $7 . 5$

Step-by-step solution

E = \dfrac{1 + cos {142}^{o}}{2} + \dfrac{1 + cos {98}^{o}}{2} + \dfrac{1}{2} \left(2cos {71}^{o}cos {49}^{o}\right)

= 1 +

\dfrac{1}{2} \left(cos {142}^{o} + cos {98}^{o}\right) +

\dfrac{1}{2}

\left(

cos {120}^{o} + cos {22}^{o}\right)

= 1 +

\dfrac{1}{2} \left(2cos {120}^{o}cos {22}^{o}\right) + \dfrac{1}{2}\left(- \dfrac{1}{2} +cos {22}^{o}\right)

<br/>

= 1 + \left(-

\dfrac{1}{2}\right)

cos {22}^{o} -

\dfrac{1}{4} + \dfrac{1}{2}

cos {22}^{o} =

\dfrac{3}{4}<br/>

E =

\dfrac{3}{4} =

0.75

<br/>

⇒ 10E = 7.5

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