Question

Find the sum of the expression $\dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+...........n$ terms

Answer 1

Hint:Multiply and divide the whole expression by $\sin \theta $ .And split $\sin \theta $ as $\sin \left( 2\theta -\theta \right)$ for first term, $\sin \left( 3\theta -2\theta \right)$ for second term, $\sin \left( 4\theta -3\theta \right)$ for third term and similarly so on. Use the trigonometry identity given as
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$.

Complete step-by-step answer:
Given series in the equation is
$\dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+...........n$
Let sum of the given series be ‘s’, we get
$s=\dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+...........\text{ }.......\left( i \right)$
Multiply and divide by $\sin \theta $ in the right-hand side of the above equation
So, we get value of s as
$s=\dfrac{\sin \theta }{\sin \theta }\left[ \dfrac{1}{\sin \theta \sin 2\theta }+\dfrac{1}{\sin 2\theta \sin 3\theta }+\dfrac{1}{\sin 3\theta \sin 4\theta }+........... \right]$
\[s=\dfrac{1}{\sin \theta }\left[ \dfrac{\sin \theta }{\sin \theta \sin 2\theta }+\dfrac{\sin \theta }{\sin 2\theta \sin 3\theta }+\dfrac{\sin \theta }{\sin 3\theta \sin 4\theta }+........... \right]\text{ }......\left( ii \right)\]
Now, we can write the last term of the given series as
$\dfrac{\sin \theta }{\sin \left( n-1 \right)\theta \sin n\theta }........\left( iii \right)$
Because, the difference in denominator terms in the product is $\theta $ .So, we can write the general term or last term as expression given in the problem
So, we get equation (ii) as
$s=\dfrac{1}{\sin \theta }\left[ \dfrac{\sin \theta }{\sin \theta \sin 2\theta }+\dfrac{\sin \theta }{\sin 2\theta \sin 3\theta }+\dfrac{\sin \theta }{\sin 3\theta \sin 4\theta }+.........\dfrac{\sin \theta }{\sin \left( n-1 \right)\theta \sin n\theta } \right]......\left( v \right)$
Now, replace $\sin \theta $ in numerator of all terms in the bracket as
$\begin{align}
  & \sin \theta =\sin \left( 2\theta -\theta \right)\text{ }{{\text{1}}^{st}}\text{ term} \\
 & \sin \theta =\sin \left( 3\theta -2\theta \right)\text{ }{{\text{2}}^{nd}}\text{ term} \\
 & \sin \theta =\sin \left( 4\theta -3\theta \right)\text{ }{{\text{3}}^{rd}}\text{ term} \\
\end{align}$
Similarly, we can write $\sin \theta $ for nth term as
$\sin \theta =\sin \left( n\theta -\left( n-1 \right)\theta \right)\text{ }{{\text{n}}^{th}}\text{ term}$
Hence, we can write equation (iv) with the help of above equations as
$s=\dfrac{1}{\sin \theta }\left[ \dfrac{\sin \left( 2\theta -\theta \right)}{\sin \theta \sin 2\theta }+\dfrac{\sin \left( 3\theta -2\theta \right)}{\sin 2\theta \sin 3\theta }+\dfrac{\sin \left( 4\theta -3\theta \right)}{\sin 3\theta \sin 4\theta }+.........\dfrac{\sin \left( \left( n\theta \right)-\left( n-1 \right)\theta \right)}{\sin \left( n-1 \right)\theta \sin n\theta } \right]$
Now use the trigonometric identity given as
$\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
Hence, we can write series S using the above identity as
\[s=\dfrac{1}{\sin \theta }\left[ \begin{align}
  & \dfrac{\sin 2\theta cos\theta -cos2\theta sin\theta }{\sin \theta \sin 2\theta }+\dfrac{\sin 3\theta cos2\theta -cos3\theta sin2\theta }{\sin 2\theta \sin 3\theta } \\
 & \text{ }+\dfrac{\sin 4\theta cos3\theta -cos4\theta sin3\theta }{\sin 3\theta \sin 4\theta }+......\dfrac{\sin \left( n\theta \right)\cos \left( n-1 \right)\theta -\sin \left( n-1 \right)\theta \cos n\theta }{\sin \left( n-1 \right)\theta \sin n\theta } \\
\end{align} \right]\]
\[s=\dfrac{1}{\sin \theta }\left[ \begin{align}
  & \dfrac{\sin 2\theta cos\theta }{\sin \theta \sin 2\theta }-\dfrac{cos2\theta sin\theta }{\sin \theta \sin 2\theta }+\dfrac{\sin 3\theta cos2\theta }{\sin 2\theta \sin 3\theta }-\dfrac{cos3\theta sin2\theta }{\sin 2\theta \sin 3\theta } \\
 & \text{ }+\dfrac{\sin 4\theta cos3\theta }{\sin 3\theta \sin 4\theta }-\dfrac{cos4\theta sin3\theta }{\sin 3\theta \sin 4\theta }+......\dfrac{\sin \left( n\theta \right)\cos \left( n-1 \right)\theta }{\sin \left( n-1 \right)\theta \sin n\theta }-\dfrac{\sin \left( n-1 \right)\theta \cos n\theta }{\sin \left( n-1 \right)\theta \sin n\theta } \\
\end{align} \right]\]
$s=\dfrac{1}{\sin \theta }\left[ \begin{align}
  & \dfrac{cos\theta }{\sin \theta }-\dfrac{cos2\theta }{\sin 2\theta }+\dfrac{cos2\theta }{\sin 2\theta }-\dfrac{cos3\theta }{\sin 3\theta }+\dfrac{cos3\theta }{\sin 3\theta }-\dfrac{cos4\theta }{\sin 4\theta } \\
 & \text{ }+.........\dfrac{\cos \left( n-1 \right)\theta }{\sin \left( n-1 \right)\theta }-\dfrac{\cos n\theta }{\sin n\theta } \\
\end{align} \right]$
We know $\dfrac{\cos x}{\sin x}=\cot x$
So, we get value of ‘s’ as
$s=\dfrac{1}{\sin \theta }\left[ \cot \theta -\cot 2\theta +\cot 2\theta -\cot 3\theta +\cot 3\theta -\cot 4\theta +.........\cot \left( n-1 \right)\theta -\operatorname{cotn}\theta \right]$
The above series can also be written as
\[s=\dfrac{1}{\sin \theta }\left[ \begin{align}
  & \cot \theta -\cot 2\theta + \\
 & \cot 2\theta -\cot 3\theta + \\
 & \cot 3\theta -\cot 4\theta + \\
 & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\
 & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\
 & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\
 & .\text{ }\text{. }\text{. }\text{. }\text{. }\text{.} \\
 & \cot \left( n-1 \right)\theta -\cot n\theta \\
\end{align} \right]\]
So, we get value of ‘s’ as
$\begin{align}
  & s=\dfrac{1}{\sin \theta }\left[ \cot \theta -\cot n\theta \right] \\
 & s=\dfrac{1}{\sin \theta }\left[ \dfrac{\cos \theta }{\sin \theta }-\dfrac{\cos n\theta }{\sin n\theta } \right] \\
 & s=\dfrac{1}{\sin \theta }\left[ \dfrac{\cos \theta \sin n\theta -\cos n\theta \sin \theta }{\sin \theta \sin n\theta } \right] \\
 & s=\dfrac{1}{\sin \theta }\dfrac{\sin \left( n\theta -\theta \right)}{\sin \theta \sin n\theta } \\
 & s=\dfrac{\sin \left( n-1 \right)\theta }{{{\sin }^{2}}\theta \sin n\theta } \\
\end{align}$
Hence, sum of given series is
$s=\dfrac{\cot \theta -\cot n\theta }{\sin \theta }=\dfrac{\sin \left( n-1 \right)\theta }{{{\sin }^{2}}\theta \sin n\theta }$

Note: If general term of any series is
${{T}_{n}}=\sum{\dfrac{1}{\cos \left( n-1 \right)\theta \sin n\theta }}$
Then divide and multiply the whole equation by $\cos \theta =\cos \left( n\theta -\left( n-1 \right)\theta \right)$
And if ${{T}_{n}}$ is of type
\[{{T}_{n}}=\sum{\dfrac{1}{\cos \left( n-1 \right)\theta \cos n\theta }\text{ (or) }}\sum{\dfrac{1}{\cos \left( n-1 \right)\theta \sin n\theta }\text{ }}\]
Then divide and multiply the whole equation by \[\sin \left( n\theta -\left( n-1 \right)\theta \right)\]
Calculation is the important side of this question as well. And creating a difference type with the help of observation is the key point of the solution.