Question

$\sf\sec \bigg( \sum \limits^{2018}_{k = - 1} \csc( \tan( \cot( \sin( \cos( \frac{4k + 5}{20} \pi ) ) ) ) ) \bigg )$
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Answer 1

Refer to the attachment!!

Answer 2

Answer:

$\huge \red{1}$

Step-by-step explanation:

$\sf\sec \bigg( \sum \limits^{2018}_{k = - 1} \csc( \tan( \cot( \sin( \cos( \frac{4k + 5}{20} \pi ) ) ) ) ) \bigg)$

Let

$f(x) = \csc( \tan( \cot( \sin( x ) ) )$

Notice that all functions Cosec , tan , cot and sine are odd functions so composite of these functions f(x) will also be an odd function . i.e.

$f( - x) = - f(x)$

Now let's take summation part

$s = \sum \limits^{2018}_{k = - 1}f( \cos( \frac{4k + 5}{20} \pi)) \\ \\ = \sum \limits^{2018}_{k = - 1}f( \cos ( \frac{k }{5} \pi + \frac{\pi}{4} ))$

Now we know that cosine is a periodic function so it's value gonna repeat after some terms of summation . Let's check

Using the fact that

$\boxed{ \cos(\pi + x )= - \cos(x) } \\ \boxed{ \cos(2\pi + x) = \cos(x) }$

First term of series k = -1

$a _{1} = f( \cos( - \frac{\pi}{5} + \frac{\pi}{4} ))$

Sixth term of series k =4

$a_{6}= f( \cos( \frac{ 4\pi}{5} + \frac{\pi}{4} )) \\ \\ = f( \cos(\pi - \frac{ \pi}{5} + \frac{\pi}{4} ) ) \\ \\ = f( - \cos( - \frac{\pi}{5} + \frac{\pi}{4} ))\\ \\=-f( \cos( - \frac{\pi}{5} + \frac{\pi}{4} )) = - a_{1}$

And 11 th term of series , k = 9

$a_{11}= f( \cos( \frac{ 9\pi}{5} + \frac{\pi}{4} ) ) \\ \\ =f( \cos(2\pi - \frac{ \pi}{5} + \frac{\pi}{4} )) \\ \\ = f(\cos( - \frac{\pi}{5} + \frac{\pi}{4} ) ) = a_{1}$

So there is a cycle of 10 terms and sum of first 10 terms will be zero because first term will cancel out by 6 th , 2nd term by 7th and so on i.e.each rth term of series will be cancelled by (r+5) th term . After 10 terms again cycle repeats .

Total number of terms in our sum ( k starts from -1 so don't forget to add -1 , 0 )

$\frac{2018 + 2}{2} = 1010 = 10 \times 101$

So n is multiple of 10 I e. sum s will be zero .

Now required answer

$= \sec(s) = \sec(0) = 1$