QJEEM21Mae8fa: JEE Main 2021 Mathematics - Class 12 JEEM21Mae8fa Solution: If 0 < a, b < 1, and tan − 1 a + tan − 1 b = π 4 , then the…

Question

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If 0 < a, b < 1, and tan^{$-$ 1}a + tan^{$-$ 1}b =

\frac{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 :

Accepted Answer

tan^{$-$ 1}a + tan^{$-$ 1}b =

\frac{π}{4}

0 < a, b < 1

\Rightarrow \frac{a + b}{1 - a b} = 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)

(

∵

expansion of log_e(1 + x))

= \log_{e} [(1 + a) (1 + b)]

= \log_{e} 2

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