Answer:
[tex]\qquad\boxed{{\sf \: \displaystyle\sum_{r=1}^n\bf u_r = \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(1)\bigg) \: }}\\ \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf \: u_r \: = \: \dfrac{1}{cos(r + 1) \: cosr} \\ \\ [/tex]
can be rewritten as
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin1}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1 - r)}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1)cosr - sinrcos(r + 1)}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: sin(x + y) = sinxcosy + cosysinx \: }} \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1)cosr}{cos(r + 1) \: cosr} - \dfrac{cos(r + 1)sinr}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1)}{cos(r + 1) \:} - \dfrac{sinr}{ \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( tan(r + 1) - tanr\bigg) \\ \\ [/tex]
So, on substituting r = 1, 2, 3, ..., n, we get
[tex]\sf \: u_1 \: = \: \dfrac{1}{sin1}\bigg( tan(2) - tan(1)\bigg) \\ \\ [/tex]
[tex]\sf \: u_2 \: = \: \dfrac{1}{sin1}\bigg( tan(3) - tan(2)\bigg) \\ \\ [/tex]
[tex]\sf \: u_3 \: = \: \dfrac{1}{sin1}\bigg( tan(4) - tan(3)\bigg) \\ \\ [/tex]
.
[tex]\sf \: u_n \: = \: \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(n)\bigg) \\ \\ [/tex]
On adding above all, we get
[tex]\sf \: u_1 + u_2 + u_3 + ... + u_n = \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(1)\bigg)\\ \\ [/tex]
Hence,
[tex]\sf\implies \: \sf \: \displaystyle\sum_{r=1}^n\bf \: u_r = \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(1)\bigg)\\ \\ [/tex]
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Answers & Comments
Answer:
[tex]\qquad\boxed{{\sf \: \displaystyle\sum_{r=1}^n\bf u_r = \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(1)\bigg) \: }}\\ \\ [/tex]
Step-by-step explanation:
Given that,
[tex]\sf \: u_r \: = \: \dfrac{1}{cos(r + 1) \: cosr} \\ \\ [/tex]
can be rewritten as
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin1}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1 - r)}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1)cosr - sinrcos(r + 1)}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: sin(x + y) = sinxcosy + cosysinx \: }} \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1)cosr}{cos(r + 1) \: cosr} - \dfrac{cos(r + 1)sinr}{cos(r + 1) \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( \dfrac{sin(r + 1)}{cos(r + 1) \:} - \dfrac{sinr}{ \: cosr}\bigg) \\ \\ [/tex]
[tex]\sf \: u_r \: = \: \dfrac{1}{sin1}\bigg( tan(r + 1) - tanr\bigg) \\ \\ [/tex]
So, on substituting r = 1, 2, 3, ..., n, we get
[tex]\sf \: u_1 \: = \: \dfrac{1}{sin1}\bigg( tan(2) - tan(1)\bigg) \\ \\ [/tex]
[tex]\sf \: u_2 \: = \: \dfrac{1}{sin1}\bigg( tan(3) - tan(2)\bigg) \\ \\ [/tex]
[tex]\sf \: u_3 \: = \: \dfrac{1}{sin1}\bigg( tan(4) - tan(3)\bigg) \\ \\ [/tex]
.
.
.
.
[tex]\sf \: u_n \: = \: \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(n)\bigg) \\ \\ [/tex]
On adding above all, we get
[tex]\sf \: u_1 + u_2 + u_3 + ... + u_n = \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(1)\bigg)\\ \\ [/tex]
Hence,
[tex]\sf\implies \: \sf \: \displaystyle\sum_{r=1}^n\bf \: u_r = \dfrac{1}{sin1}\bigg( tan(n + 1) - tan(1)\bigg)\\ \\ [/tex]