If 0 < a, b < 1, and tan-1a + tan-1b =π/4, then the value of (a - b)-((a^2+b^2)/2)+((a^3+b^3)/3)-((a^4+b^4)/4)+.... is

Question

If 0 < a, b < 1, and tan^-1a + tan^-1b =π/4, then the value of

\((a+b)-(\frac{a^2+b^2}2)+(\frac{a^3+b^3}3)-(\frac{a^4+b^4}4)+....\) is

(1) \(log_e2\)

(2) \(e^2-1\)

(3) e

(4) \(log_e(\frac e2)\)

Answer 1

Correct option is (1) \(log_e2\)

tan^-1a + tan^-1b = π/4    0 < a, b < 1

⇒ \(\frac{a+b}{1-ab}=1\)

a + b = 1 - ab

(a + 1)(b + 1) = 2

Now \([a-\frac{a^2}2+\frac{a^3}3+....]\)\(+[b-\frac{b^2}2+\frac{b^3}3+...]\)

\(=log_e(1+a)+log_e(1+b)\)

(\(\because\) expansion of log_e(1 + x))

= log_e[(1 + a)(1 + x)]

= log_e2