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Question: If \(\tan \theta = \sqrt {\dfrac{3}{2}} \) ,then the sum of the infinite series \(1 + 2(1 - \cos \th...

If tan⁡θ=32\tan \theta = \sqrt {\dfrac{3}{2}} ,then the sum of the infinite series 1+2(1−cos⁡θ)+3(1−cos⁡θ)2+4(1−cos⁡θ)3+...∞1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty is
A) 23\dfrac{2}{3}
B) 34\dfrac{{\sqrt 3 }}{4}
C) 522\dfrac{5}{{2\sqrt 2 }}
D) 52\dfrac{5}{2}

Explanation

Solution

In the above question we have been given the value of tan⁡θ\tan \theta . We will first assume the given infinite series as
SS i.e. S=1+2(1−cos⁡θ)+3(1−cos⁡θ)2+4(1−cos⁡θ)3+...∞S = 1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty .
After that we will multiply the expression with (1−cos⁡θ)(1 - \cos \theta ) .
The general form of the infinite geometric series is a,ar2,ar3...arna,a{r^2},a{r^3}...a{r^n} .
By comparing the given series in the question, we can see that the value of r=(1−cos⁡θ)r = (1 - \cos \theta ) .
We will apply the formula of sum of infinites series i.e.
S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}} , where aa is the first term i.e. 11
and rr is the common ratio, (1−cos⁡θ)(1 - \cos \theta ) .

Complete step by step answer:
We have been given the infinite geometric series 1+2(1−cos⁡θ)+3(1−cos⁡θ)2+4(1−cos⁡θ)3+...∞1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty .
Let us assume that S=1+2(1−cos⁡θ)+3(1−cos⁡θ)2+4(1−cos⁡θ)3+...∞S = 1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty .
This is our first equation.
We will multiply both the side of the equation with (1−cos⁡θ)(1 - \cos \theta )
S(1 - \cos \theta ) = (1 - \cos \theta )\left\\{ {1 + 2(1 - \cos \theta ) + 3{{(1 - \cos \theta )}^2} + 4{{(1 - \cos \theta )}^3} + ...\infty } \right\\} .
We can write the above also as S - S\cos \theta = \left\\{ {1(1 - \cos \theta ) + 2(1 - \cos \theta )(1 - \cos \theta ) + 3{{(1 - \cos \theta )}^2}(1 - \cos \theta ) + 4{{(1 - \cos \theta )}^3}(1 - \cos \theta ) + ...\infty } \right\\}On multiplying the above expression can be written as S - S\cos \theta = \left\\{ {(1 - \cos \theta ) + 2{{(1 - \cos \theta )}^2} + 3{{(1 - \cos \theta )}^3} + 4{{(1 - \cos \theta )}^4} + ...\infty } \right\\}
This is our second equation.
Now we will subtract second equation from first equation i.e.
S - \left( {S - S\cos \theta } \right) = \left\\{ {1 + 2(1 - \cos \theta ) + 3{{(1 - \cos \theta )}^2} + 4{{(1 - \cos \theta )}^3} + ...\infty } \right\\} - \left\\{ {(1 - \cos \theta ) + 2{{(1 - \cos \theta )}^2} + 3{{(1 - \cos \theta )}^3} + 4{{(1 - \cos \theta )}^4} + ...\infty } \right\\}By breaking the values we can write: S−S+Scos⁡θ=1+2(1−cos⁡θ)−(1−cos⁡θ)+3(1−cos⁡θ)2−2(1−cos⁡θ)2+...∞S - S + S\cos \theta = 1 + 2(1 - \cos \theta ) - (1 - \cos \theta ) + 3{(1 - \cos \theta )^2} - 2{(1 - \cos \theta )^2} + ...\infty
On subtracting the values above we have
Scos⁡θ=1+(1−cos⁡θ)+(1−cos⁡θ)2+...∞S\cos \theta = 1 + (1 - \cos \theta ) + {(1 - \cos \theta )^2} + ...\infty
Now here we have the infinite series. So we will apply the formula of sum of the infinite series i.e.
S∞=a1−r{S_\infty } = \dfrac{a}{{1 - r}} , where aa is the first term i.e.
11 and,
rr is the common ratio,
(1−cos⁡θ)(1 - \cos \theta )
By putting the values in the formula we have
Scos⁡θ=11−(1−cos⁡θ)S\cos \theta = \dfrac{1}{{1 - (1 - \cos \theta )}}
We can break down the value and it gives us
11−1+cos⁡θ=1cos⁡θ\dfrac{1}{{1 - 1 + \cos \theta }} = \dfrac{1}{{\cos \theta }}
By can transfer the cosine function to the other side i.e.
S=1cos⁡θ×cos⁡θ=1cos⁡2θS = \dfrac{1}{{\cos \theta \times \cos \theta }} = \dfrac{1}{{{{\cos }^2}\theta }}
We know that
1cos⁡θ=sec⁡θ\dfrac{1}{{\cos \theta }} = \sec \theta
So by putting this we can write
S=sec⁡2θS = {\sec ^2}\theta
WE know the trigonometric identity that
sec⁡2θ=1+tan⁡2θ{\sec ^2}\theta = 1 + {\tan ^2}\theta , we will put this in the expression
We have,
S=1+tan⁡2θS = 1 + {\tan ^2}\theta ,
And the value of tan⁡θ\tan \theta is given, so we will put that in the expression:
S=1+(32)2S = 1 + {\left( {\sqrt {\dfrac{3}{2}} } \right)^2}
On simplifying we have
1+32=2+321 + \dfrac{3}{2} = \dfrac{{2 + 3}}{2}
It gives us the value
S=52S = \dfrac{5}{2}
Therefore, If tan⁡θ=32\tan \theta = \sqrt {\dfrac{3}{2}} ,then the sum of the infinite series 1+2(1−cos⁡θ)+3(1−cos⁡θ)2+4(1−cos⁡θ)3+...∞1 + 2(1 - \cos \theta ) + 3{(1 - \cos \theta )^2} + 4{(1 - \cos \theta )^3} + ...\infty is 52\dfrac{5}{2}. Hence the correct option is (D).

Note:
We should know that in the above we have been given to find sum of infinite series, i.e.
here nn goes to ∞\infty
If we have to find the sum of finite geometric series is
Sn=a(1−rn1−r){S_n} = a\left( {\dfrac{{1 - {r^n}}}{{1 - r}}} \right) , where nn is the number of terms.