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Kandarp's Observation

If \( a_1, a_2, a_3, \ldots a_n \) are in an Arithmetic Progression with common difference \( d \neq 0 \), then prove that

\[ \sin (d) \times ( \csc (a_1) \times \csc (a_2) + \csc (a_2) \times \csc (a_3) + \ldots + \csc (a_{n-1}) \times \csc (a_n) ) = \cot( a_ 1 )- \cot (a_n.) \]

Note by Calvin Lin
3 years ago

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First open all the terms to get \(LHS=sin(d)csc({ a }_{ 1 })csc({ a }_{ 2 })+sin(d)csc({ a }_{ 2 })csc({ a }_{ 3 })+................sin(d)csc({ a }_{ n-1 })csc({ a }_{ n })\\ =\frac { sin(d) }{ sin({ a }_{ 1 })sin({ a }_{ 2 }) } +\frac { sin(d) }{ sin({ a }_{ 2 })sin({ a }_{ 3 }) } +............+\frac { sin(d) }{ sin({ a }_{ n-1 })sin({ a }_{ n }) } .\)

Now since \({ a }_{ 1 },{ a }_{ 2 },......{ a }_{ n }\) are in \(A.P.\)

\(d={ a }_{ 2 }-{ a }_{ 1 }={ a }_{ 3 }-{ a }_{ 2 }=............={ a }_{ n }-{ a }_{ n-1 }.\)

Now replacing \(d\) suitably we get

\(LHS=\frac { sin({ a }_{ 2 }-{ a }_{ 1 }) }{ sin({ a }_{ 1 })sin({ a }_{ 2 }) } +\frac { sin({ a }_{ 3 }-{ a }_{ 2 }) }{ sin({ a }_{ 2 })sin({ a }_{ 3 }) } +............+\frac { sin({ a }_{ n }-{ a }_{ n-1 }) }{ sin({ a }_{ n-1 })sin({ a }_{ n }) } .\)

Using identity \(sin(A-B)=sin(A)cos(B)-sin(B)cos(A)\) we get

\(\frac { sin(A-B) }{ sin(A)sin(B) } =\frac { sin(A)cos(B)-sin(B)cos(A) }{ sin(A)sin(B) } =cot(B)-cot(A).\)

Now our \(LHS\) becomes \(=\quad (cot({ a }_{ 2 })-cot({ a }_{ 1 }))+(cot({ a }_{ 3 })-cot({ a }_{ 2 }))+...........+(cot({ a }_{ n })-cot({ a }_{ n-1 }))\)

Note how terms cancel here hence \(LHS\) becomes \(=\quad cot({ a }_{ n })-cot({ a }_{ 1 })=RHS\quad \)

\(Hence\quad proved\) Ronak Agarwal · 3 years ago

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@Kandarp Singh I believe that you will get better response posting this as a note for discussion, instead of as a question looking for a numerical answer. Calvin Lin Staff · 3 years ago

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@Calvin Lin ok Kandarp Singh · 3 years ago

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The \(RHS\) kinda suggests that this is telescopical. Indeed, \(sin(d)*csc(a_n)*csc(a_{n+1})=\frac {sin(a_{n+1}-a_{n})}{sin(a_n)*sin(a_{n+1})}\) Using the difference formula, it simplifies to \(cot(a_n)-cot(a_{n+1})\).

Hence LHS=\(cot(a_1)-cot(a_2)+cot(a_2)-cot(a_3)+...+cot(a_{n-1})-cot{a_n}=cot(a_1)-cot(a_n)\) Xuming Liang · 3 years ago

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