Question

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}\] are in A.P with common difference \[d\left( d\ne 0 \right)\] then prove that
\[\sin d\left( \csc {{a}_{1}}\csc {{a}_{2}}+\csc {{a}_{2}}\csc {{a}_{3}}+....+\csc {{a}_{n-1}}\csc {{a}_{n}} \right)=\cot {{a}_{1}}-\cot {{a}_{n}}\]

Answer 1

Hint: We solve this problem by using the composite angle formula and common difference of an A.P
We use the condition that the common difference of an A.P is given as the difference of any two consecutive terms in the A.P that is if \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}\] are in A.P with common difference \[d\] then
\[\Rightarrow d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=....={{a}_{n}}-{{a}_{n-1}}\]
We have the formula of composite angle for sine function as
\[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\]


Complete step by step answer:
We are given that \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}\] are in A.P with common difference \[d\left( d\ne 0 \right)\]
We are asked to prove that
\[\sin d\left( \csc {{a}_{1}}\csc {{a}_{2}}+\csc {{a}_{2}}\csc {{a}_{3}}+....+\csc {{a}_{n-1}}\csc {{a}_{n}} \right)=\cot {{a}_{1}}-\cot {{a}_{n}}\]
Let us assume that the LHS of given result as
\[\Rightarrow L=\sin d\left( \csc {{a}_{1}}\csc {{a}_{2}}+\csc {{a}_{2}}\csc {{a}_{3}}+....+\csc {{a}_{n-1}}\csc {{a}_{n}} \right)\]
We know that the formula of cosecant function as
\[\csc \theta =\dfrac{1}{\sin \theta }\]
By using the above formula and multiplying the outside term to inside the bracket in LHS we get
\[\begin{align}
  & \Rightarrow L=\sin d\left( \dfrac{1}{\sin {{a}_{1}}\sin {{a}_{2}}}+\dfrac{1}{\sin {{a}_{2}}\sin {{a}_{3}}}+...+\dfrac{1}{\sin {{a}_{n-1}}\sin {{a}_{n}}} \right) \\
 & \Rightarrow L=\dfrac{\sin d}{\sin {{a}_{1}}\sin {{a}_{2}}}+\dfrac{\sin d}{\sin {{a}_{2}}\sin {{a}_{3}}}+...+\dfrac{\sin d}{\sin {{a}_{n-1}}\sin {{a}_{n}}} \\
\end{align}\]
We know that the condition that the common difference of an A.P is given as the difference of any two consecutive terms in the A.P that is if \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}\] are in A.P with common difference \[d\] then
\[\Rightarrow d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=....={{a}_{n}}-{{a}_{n-1}}\]
Now, let us use the required values in the required terms of the above equation then we get
\[\Rightarrow L=\dfrac{\sin \left( {{a}_{2}}-{{a}_{1}} \right)}{\sin {{a}_{1}}\sin {{a}_{2}}}+\dfrac{\sin \left( {{a}_{3}}-{{a}_{2}} \right)}{\sin {{a}_{2}}\sin {{a}_{3}}}+...+\dfrac{\sin \left( {{a}_{n}}-{{a}_{n-1}} \right)}{\sin {{a}_{n-1}}\sin {{a}_{n}}}\]
We know that the formula of composite angle for sine function as
\[\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B\]
By using this formula of composite angle to the above equation we get

\[\Rightarrow L=\dfrac{\sin {{a}_{2}}\cos {{a}_{1}}-\cos {{a}_{2}}\sin {{a}_{1}}}{\sin {{a}_{1}}\sin {{a}_{2}}}+\dfrac{\sin {{a}_{3}}\cos {{a}_{2}}-\cos {{a}_{3}}\sin {{a}_{2}}}{\sin {{a}_{2}}\sin {{a}_{2}}}+...+\dfrac{\sin {{a}_{n}}\cos {{a}_{n-1}}-\cos {{a}_{n}}\sin {{a}_{n-1}}}{\sin {{a}_{n}}\sin {{a}_{n-1}}}\]
Now, let us divide the each term to individual terms then we get
\[\begin{align}
  & \Rightarrow L=\cot {{a}_{1}}-\cot {{a}_{2}}+\cot {{a}_{2}}-\cot {{a}_{3}}+\cot {{a}_{3}}-\cot {{a}_{4}}+...+\cot {{a}_{n-1}}-\cot {{a}_{n}} \\
 & \Rightarrow L=\cot {{a}_{1}}-\cot {{a}_{n}} \\
\end{align}\]
Here we can see that the value of LHS is equal to RHS
Therefore we can conclude that the required result has been proved that is
\[\therefore \sin d\left( \csc {{a}_{1}}\csc {{a}_{2}}+\csc {{a}_{2}}\csc {{a}_{3}}+....+\csc {{a}_{n-1}}\csc {{a}_{n}} \right)=\cot {{a}_{1}}-\cot {{a}_{n}}\]

Note:
 Students may do mistake in taking the common difference of an A.P
We know that if \[{{a}_{1}},{{a}_{2}},{{a}_{3}},....,{{a}_{n}}\] are in A.P with common difference \[d\] then
\[\Rightarrow d={{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}=....={{a}_{n}}-{{a}_{n-1}}\]
Here we can see that the difference of any two consecutive terms gives the common difference. But students may misunderstand the definition of A.P and take the common difference as the difference of only first two terms that is
\[\Rightarrow d={{a}_{2}}-{{a}_{1}}\]