Question

If a₁, a₂, a₃,...,a_n 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"))]`

Answer 1

If a₁, a₂, a₃, ..., a_n 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 a_n – tan^–1 a_{n – 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 a_n – tan^–1 a_{n – 1}]

⇒ tan [tan^–1 a_n – 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]