Primes can make answers Creative @Brilliant

$2sin2^{\circ} + 4sin4^{\circ} + 6sin6^{\circ} + ....................... + 178sin178^{\circ} + 180sin180^{\circ}$

$2sin2^{\circ} + 4sin4^{\circ} + 6sin6^{\circ} + .......................... + 178sin2^{\circ} + 0$

adding or pairing last and the first term , similarly second and second last till $88sin88^{\circ}$ , and we will not pair $90sin90^{\circ}$ .

thus we get

$180sin2^{\circ} + 180sin4^{\circ} + 180sin6^{\circ} + ....................... + 180sin88^{\circ} + 90$

$180( sin2^{\circ} + sin4^{\circ} + ..................... + sin88^{\circ}) + 90$

multiplying and dividing by $2sin1^{\circ}$

$\frac{180}{2sin1^{\circ}}( 2sin1^{\circ}sin2^{\circ} + 2sin1^{\circ}sin4^{\circ} + ............... + 2sin1^{\circ}sin88^{\circ}) + 90$

now $2sina^{\circ} sinb^{\circ} = cos(a - b) - cos(a + b)$

thus,

$\frac{180}{2sin1^{\circ}}(cos1^{\circ} - cos3^{\circ} + cos3^{\circ} - cos5^{\circ} + cos5^{\circ} ....... - cos89^{\circ} ) + 90$

$\frac{180}{2sin1^{\circ}}(cos1^{\circ} - sin(90 - 89)^{\circ}) + 90$

$90cot1^{\circ} - 90 + 90$

$= 90cot1^{\circ}$

Anyone is having other method?

$2sin2 + 4sin4 + 6sin6 +..... + 180sin180$

= $2(sin2 + 2sin4 + 3sin6 + ..... + 90sin180$

Now, $sin2 + 2 sin4 + 3sin6 +.... + 90sin180 = (sin2 + sin4 + sin6 +... sin180) + (sin4 + sin6 +.... sin 180) + (sin6 +....sin180) +...... (sin178 + sin180) + (sin180)$

Hence there are 90 such sums of sines with angles in AP and different first term.

Using the formula $sin\alpha + sin(\alpha + \beta) +..... sin(\alpha + (n-1)\beta) = \frac{sin\frac{n\beta}{2}}{sin\frac{\beta}{2}} sin(\alpha + (n-1)\frac{\beta}{2})$ repeatedly, we get

$sin2 + 2 sin4 + 3sin6 +.... + 90sin180 = (sin2 + sin4 + sin6 +... sin180) + (sin4 + sin6 +.... sin 180) + (sin6 +....sin180) +...... (sin178 + sin180) + (sin180) = \frac{sin90}{sin1} sin91 + \frac{sin89}{sin1} sin 92 + ..... + \frac{sin1}{sin1} sin 180$

= $\frac{(sin90)(sin91) + (sin89)(sin92) + .... + (sin1)(sin180)}{sin1}$

Now, $sin(90 + x) = cos x$

= $\frac{(sin90)(cos1) + (sin89)(cos2) +.... + (sin2)(cos89) + (sin1)(cos90)}{sin1}$

Now, \(sinAcosB + sinBcosA = sin(A+B)

= \(\frac{(sin90cos1 + cos90sin1) + (sin89cos2 + sin2cos89) + ..... (sin46cos45 + sin45cos46)}{sin1}\)

= $\frac{sin91 + sin91 + sin91 +..... sin91}{sin1}$

= $\frac{45*sin91}{sin1}$

As we have already seen that sin91 = cos1,

= $45*cot 1$

Now, we have already taken a 2 out of the product, hence, the correct answer is

$\boxed{90 * cot 1}$

This is the same problem as USAMO 1996 p1

we know that

sin x = sin(180-x)

we will be using this formula here.

2sin2 +4sin4+...+180 sin180

=2sin2 +4sin4 +..178 sin 178+0...as sin 180 =0

now since sin2 = sin 178, sin4 = sin176...

we get

given expression

=(2sin2 +178 sin2)+(4 sin 4 + 176 sin4) +....+ (88sin88+92 sin88)+90sin 90

=180 sin 2+180 sin4+...+180 sin 88++90

=180 (sin2 +sin4 +...+sin88)+90

=90[(2sin1 sin2+2sin1 sin4+...+2sin1 sin88)/sin1]+90

=90{[(cos1-cos3)+(cos3-cos5)+...+(cos87-cos89)]}/2+{90}

=90[(cos1-cos89)/sin1]+90

=90[(cos1/sin1)-(cos89/sin1)]+90

but cos89=sin1

so

given expression

=90[cot1-(sin1/sin1)]+90

=90 cot1 -90+90

=90 cot 1