Question

$\displaystyle \sum \limits_{k = 1}^{2020} \frac{1}{ \sin(k) \degree \sin(k + 1) \degree }$

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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle \sum \limits_{k = 1}^{2020} \frac{1}{ \sin(k) \degree \sin(k + 1) \degree }$

can be rewritten as

$\rm \:  =  \:\dfrac{1}{sin1\degree } \displaystyle \sum \limits_{k = 1}^{2020} \frac{sin1\degree }{ \sin(k) \degree \sin(k + 1) \degree }$

$\rm \:  =  \:\dfrac{1}{sin1\degree } \displaystyle \sum \limits_{k = 1}^{2020} \frac{sin(k + 1 - k)\degree }{ \sin(k) \degree \sin(k + 1) \degree }$

We know,

$\boxed{\tt{ sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: }}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{sin1\degree }\displaystyle \sum \limits_{k = 1}^{2020} \frac{sin(k + 1)\degree cos(k)\degree - sin(k)\degree cos(k + 1)\degree }{ \sin(k) \degree \sin(k + 1) \degree }$

$\rm \:  =  \: \dfrac{1}{sin1\degree }\displaystyle \sum \limits_{k = 1}^{2020} \bigg[\frac{sin(k + 1)\degree cos(k)\degree}{ \sin(k) \degree \sin(k + 1) \degree } - \dfrac{ sin(k)\degree cos(k + 1)\degree }{sin(k)\degree sin(k + 1)\degree }\bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree }\displaystyle \sum \limits_{k = 1}^{2020} \bigg[tan(k + 1)\degree - tan(k)\degree \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[(tan(2)\degree - tan(1)\degree) + (tan(3)\degree - tan(2)\degree ) + - - - + (tan(2021)\degree - tan(2020)\degree ) \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[tan(2021)\degree - tan(1)\degree\bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin2021\degree }{cos2021\degree } - \dfrac{sin1\degree }{cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin2021\degree \: cos1\degree - sin1\degree \: cos2021\degree }{cos2021\degree \: cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin(2021 \: - 1)\degree }{cos2021\degree \: cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin(2020)\degree }{cos2021\degree \: cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{sin(2020)\degree }{sin(1)\degree \: cos(1)\degree \: cos(2021)\degree }$

Hence,

$\boxed{\tt{ \rm \:\displaystyle \sum \limits_{k = 1}^{2020} \frac{1}{ \sin(k) \degree \sin(k + 1) \degree }  =  \: \dfrac{sin(2020)\degree }{sin(1)\degree \: cos(1)\degree \: cos(2021)\degree } }}$

Answer 2

Answer:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle \sum \limits_{k = 1}^{2020} \frac{1}{ \sin(k) \degree \sin(k + 1) \degree }$

can be rewritten as

$\rm \:  =  \:\dfrac{1}{sin1\degree } \displaystyle \sum \limits_{k = 1}^{2020} \frac{sin1\degree }{ \sin(k) \degree \sin(k + 1) \degree }$

$\rm \:  =  \:\dfrac{1}{sin1\degree } \displaystyle \sum \limits_{k = 1}^{2020} \frac{sin(k + 1 - k)\degree }{ \sin(k) \degree \sin(k + 1) \degree }$

We know,

$\boxed{\tt{ sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: }}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{sin1\degree }\displaystyle \sum \limits_{k = 1}^{2020} \frac{sin(k + 1)\degree cos(k)\degree - sin(k)\degree cos(k + 1)\degree }{ \sin(k) \degree \sin(k + 1) \degree }$

$\rm \:  =  \: \dfrac{1}{sin1\degree }\displaystyle \sum \limits_{k = 1}^{2020} \bigg[\frac{sin(k + 1)\degree cos(k)\degree}{ \sin(k) \degree \sin(k + 1) \degree } - \dfrac{ sin(k)\degree cos(k + 1)\degree }{sin(k)\degree sin(k + 1)\degree }\bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree }\displaystyle \sum \limits_{k = 1}^{2020} \bigg[tan(k + 1)\degree - tan(k)\degree \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[(tan(2)\degree - tan(1)\degree) + (tan(3)\degree - tan(2)\degree ) + - - - + (tan(2021)\degree - tan(2020)\degree ) \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[tan(2021)\degree - tan(1)\degree\bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin2021\degree }{cos2021\degree } - \dfrac{sin1\degree }{cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin2021\degree \: cos1\degree - sin1\degree \: cos2021\degree }{cos2021\degree \: cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin(2021 \: - 1)\degree }{cos2021\degree \: cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{1}{sin1\degree } \bigg[\dfrac{sin(2020)\degree }{cos2021\degree \: cos1\degree } \bigg]$

$\rm \:  =  \: \dfrac{sin(2020)\degree }{sin(1)\degree \: cos(1)\degree \: cos(2021)\degree }$

Hence,

$\boxed{\tt{ \rm \:\displaystyle \sum \limits_{k = 1}^{2020} \frac{1}{ \sin(k) \degree \sin(k + 1) \degree }  =  \: \dfrac{sin(2020)\degree }{sin(1)\degree \: cos(1)\degree \: cos(2021)\degree } }}$

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