The value of the expression (2 ( 1 ° + 2 ° + 3 ° + . . . . . . . . + 89 ° ))/(2 ( 1 ° + 2 ° + 3 ° + . . . . . . . . + 44 ° ) + 1) is equal to

Correct answer: A. 2 2\ (sin 1 ^ o + sin 89 ^ o ) + (sin 2 ^ o + sin 88 ^ o ) + . . . . . . . + (sin 44 ^ o + sin 46 ^ o ) + sin 45 ^ o \ 2(cos 1 ^ o + cos 2 ^ o + cos 3 ^ o + . . . . . . . + cos 44 ^ o ) + 1 = 2\ 2sin 45 ^ o (cos 44 ^ o + cos 43 ^ o + . . . . . . . + cos 1 ^ o ) + sin 45 ^ o \ 2(cos 1 ^ o + cos 2 ^ o + cos 3 ^ o + . . . . . . . + cos 44 ^ o ) + 1 = 2sin 45 ^ o \ 2(cos 1 ^ o + cos 2 ^ o + . . . . . . . + cos 44 ^ o ) + 1\ 2(cos 1 ^ o + cos 2 ^ o + . . . . . . . + cos 44 ^ o ) + 1 = 2 ⋅ 1 2 = 2

JEE Main

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

JEE Main · Mathematics · Trigonometric Ratios & Identities

The value of the expression (2 ( 1 ° + 2 ° + 3 ° + . . . . . . . ....

The value of the expression

\dfrac{2 ( \sin 1 ° + \sin 2 ° + \sin 3 ° + . . . . . . . . + \sin 89 ° )}{2 ( \cos 1 ° + \cos 2 ° + \cos 3 ° + . . . . . . . . + \cos 44 ° ) + 1}

is equal to

Options

A $\sqrt{2}$
B $\dfrac{1}{\sqrt{2}}$
C $\dfrac{1}{2}$
D0

Correct answer

A. $\sqrt{2}$

Step-by-step solution

\dfrac{2\left\{\left(sin{1}^{o} + sin{89}^{o}\right) + \left(sin{2}^{o} + sin{88}^{o}\right) + . . . . . . . + \left(sin{44}^{o} + sin{46}^{o}\right) + sin{45}^{o}\right\}}{2\left(cos{1}^{o} + cos{2}^{o} + cos{3}^{o} + . . . . . . . + cos{44}^{o}\right) + 1}

= \dfrac{2\left\{2sin{45}^{o}\left(cos{44}^{o} + cos{43}^{o} + . . . . . . . + cos{1}^{o}\right) + sin{45}^{o}\right\}}{2\left(cos{1}^{o} + cos{2}^{o} + cos{3}^{o} + . . . . . . . + cos{44}^{o}\right) + 1}

= \dfrac{2sin{45}^{o}\left\{2\left(cos{1}^{o} + cos{2}^{o} + . . . . . . . + cos{44}^{o}\right) + 1\right\}}{2\left(cos{1}^{o} + cos{2}^{o} + . . . . . . . + cos{44}^{o}\right) + 1}

= 2 ⋅ \dfrac{1}{\sqrt{2}}

= \sqrt{2}

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