If $a_{1}, a_{2}, \ldots, a_{n}$ are in A.P. with common difference $d$ (where $d \neq 0$ ); then the sum of the series $\sin d\left(\operatorname{cosec} a_{1} \operatorname{cosec} a_{2}+\operatorname{cosec} a_{2} \operatorname{cosec} a_{3}+\ldots+\operatorname{cosec}\right.$ $\left.a_{n-1} \operatorname{cosec} a_{n}\right)$ is equal to $\cot a_{1}-\cot a_{n}$

Step 1: Analyze the general term

The given series involves terms of the form $\operatorname{cosec} a_k \operatorname{cosec} a_{k+1}$. We can rewrite this using the definition of cosecant as $1/(\sin a_k \sin a_{k+1})$. This will help us simplify the expression.

Step 2: Introduce $\sin d$ into the general term

Since $a_1, a_2, \ldots, a_n$ are in A.P. with common difference $d$, we know that $a_{k+1} - a_k = d$. We can substitute $\sin d$ with $\sin(a_{k+1} - a_k)$ in the numerator. This is a key step to apply a trigonometric identity.

Step 3: Apply trigonometric identity

Using the trigonometric identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we expand the numerator. Then, we divide each term by the denominator $\sin a_k \sin a_{k+1}$. This simplifies to $\cot a_k - \cot a_{k+1}$, which is a telescoping form.

Step 4: Sum the series

Now we sum the series. Each term is of the form $(\cot a_k - \cot a_{k+1})$. This is a telescoping series, meaning that intermediate terms cancel out. The sum reduces to the first term of the first pair and the last term of the last pair.

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