Question:
If a1,a2,a3.....an is an AP with common difference d, then evaluate the following expression : tan ( tan-1(d/(1+a1a2))+tan-1(d/(1+a2a3))+tan-1(d/(1+a3a4))+........tan-1(d/(1+a(n-1)*an)))
Answer:
Given that a1 , a2 , a3 ,..................., an-1 , an is in AP
Again d is common difference.
So d = a2 - a1 = a3 - a2 = ............... = an - an-1
Now tan[ tan-1 {d/(1+a1 *a2 )}+tan-1 {d/(1+a2 *a3 )}+tan-1 {d/(1+a3 *a4 )}+........tan-1 {d/(1+an-1 *an )}]
=> tan[ tan-1 {((a2 - a1 )/(1+a1 *a2 )}+tan-1 {(a3 - a2 )/(1+a2 *a3 )}+tan-1 {(a4 - a3 )/(1+a3 *a4 )}+........tan-1 {(an - an-1 )/(1+an-1 *an )}]
=> tan{ tan-1 (a2 ) - tan-1 (a1 ) + tan-1 (a3 ) - tan-1 (a2 ) + tan-1 (a4 ) - tan-1 (a3 ) +...........................+ tan-1 (an ) - tan-1 (an-1 )} [since tan-1 a - tan-1 b = tan-1 {(a - b )/(1+a*b)}]
=> tan{ tan-1 (an ) - tan-1 (a1 )}
=> tan{ tan-1 {(an - a1 )/(1 + an *a1 )}
=> (an - a1 )/(1 + an *a1 ) [since tan{tan-1 (x)} = x ]
So tan[ tan-1 {d/(1+a1 *a2 )}+tan-1 {d/(1+a2 *a3 )}+tan-1 {d/(1+a3 *a4 )}+........tan-1 {d/(1+an-1 *an )}] = (an - a1 )/(1 + an *a1 )
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