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If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression. daadaadaadanantan[tan-1(d1+a1a2)+tan-1(d21+a2a3)+tan-1(d1+a3a4)+...+tan-1(d1+an - Mathematics

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Question

If a1, a2, a3,...,an is an arithmetic progression with common difference d, then evaluate the following expression.

`tan[tan^-1("d"/(1 + "a"_1 "a"_2)) + tan^-1("d"/(21 + "a"_2 "a"_3)) + tan^-1("d"/(1 + "a"_3 "a"_4)) + ... + tan^-1("d"/(1 + "a"_("n" - 1) "a""n"))]`

Sum
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Solution

If a1, a2, a3, ..., an are the terms of an arithmetic progression

∴ d = a2 – a1

= a3 – a2

= a4 – a3 ....

∴ `tan[tan^-1 (("a"_2 - "a"_1)/(1 + "a"_1"a"_2)) + tan^-1 (("a"_3 - "a"_2)/(1 + "a"_2 "a"_3)) + tan^-1 (("a"_4 - "a"_3)/(1 + "a"_3 "a"_4)) + ...... + tan^-1  (("a"_"n" - "a"_("n" - 1))/(1 + "a"_("n" - 1) * "a"_"n"))]``

⇒ 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)]  .....`[because tan^-1  (x - y)/(1 + xy) = tan^-1x - tan^-1y]`

⇒ 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]

⇒ tan [tan–1 an – tan–1 a1]

⇒ `tan[tan^-1 (("a"_"n" - "a"_1)/(1 + "a"_1"a"_"n"))]`

⇒ `("a"_"n" - "a"_1)/(1 + "a"_1"a"_"n")`  .....[∵ tan (tan–1x) = x]

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Chapter 2: Inverse Trigonometric Functions - Exercise [Page 37]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 2 Inverse Trigonometric Functions
Exercise | Q 19 | Page 37

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