The average value of $sin 2 °, sin 4 °, sin 6 °, . . ., sin 180 ° is$

\frac{1}{90} cot 1 °

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\frac{1}{90} tan 1 °

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\frac{1}{90} tan 89 °

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\frac{1}{90} cot 89 °

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Solution

The correct options are
A
$\frac{1}{90} cot 1 °$
C
$\frac{1}{90} tan 89 °$
Here, we represent the sum of sines as:
$S = sin 2 ° + sin 4 ° + . . + sin 180 °$ .
Now, using the expression for sum of sines
$S = sin (α) + sin (α + β) + . . + sin (α + (n - 1) β) \Rightarrow S = \frac{sin (\frac{nβ}{2})}{sin (\frac{β}{2})} sin (α + (n - 1) \frac{β}{2}) .$
Now, for our series, we have $α = 2 °, β = 2 ° and n = 90$ .
So, we write sum as:
$S = \frac{sin (\frac{90 \times 2 °}{2})}{sin (\frac{2 °}{2})} sin (2 ° + (90 - 1) \frac{2 °}{2}) \Rightarrow S = \frac{sin (90 °)}{sin (1 °)} sin (\frac{180 ° + 2 °}{2}) \Rightarrow S = \frac{sin (90 °)}{sin (1 °)} sin (91 °) \Rightarrow S = \frac{sin (90 °)}{sin (1 °)} sin (91 °) \Rightarrow S = \frac{1}{sin (1 °)} sin (90 ° + 1 °) \Rightarrow S = \frac{cos (1 °)}{sin (1 °)} = cot (1 °) .$
Now, we have 90 terms so average of sum is given as:
$A = \frac{S}{90} = \frac{1}{90} cot (1 °)$
Also, $cot (1 °) = cot (90 ° - 89 °) = tan (89 °) \Rightarrow A = \frac{1}{90} cot 1 ° = \frac{1}{90} tan 89 ° .$

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