Let {f}: {R} \\to{R} be defined as{f}({x}+{y})+{f}({x}-{y})=2 {f}({x}) {f}({y}), {f}(\frac{1}{2})=-1 . Then the value of \sum_{k=1} ^{20} \frac{1}{\sin (k) \sin (k+f(k))} is equal to: [27 Jul 2021 Shift 2]

Correct Answer is : cosec2(1)cosec(21)sin(20)

{\cosec}^{2}(21) \cos (20) \cos

\sec ^{2}(1) \sec (21) \cos (20)

{\cosec}^{2}(1) {\cosec}(21) \sin (20)

\sec ^{2}(21) \sin (20) \sin (2)

JEE Main 27 July 2021 Shift 2 Solved Paper

Section: Mathematics

Question : 64 of 90

Marks: +1, -0

Let f:R→R be defined asf(x+y)+f(x−y)=2f(x)f(y),f(

)=−1. Then the value of

∑

k=1

sin(k)‌sin(k+f(k))

is equal to:

[27 Jul 2021 Shift 2]

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