If $0<a, b<1$, and $\tan ^{-1} a+\tan ^{-1} b=\frac{\pi}{4}$, then the value of $(a+b)-\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{3}+b^{3}}{3}\right)-\left(\frac{a^{4}+b^{4}}{4}\right)+\ldots \ldots \ldots \ldots \ldots$ is

(D) $\log _{\mathrm{e}}\left(\frac{\mathrm{e}}{2}\right)$

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Inverse Trigonometric Functions

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}) + \dots\dots\dots\dots\dots

Answer: (A)

For

∣ t ∣ < 1

t - \frac{t ^{2}}{2} + \frac{t ^{3}}{3} - \dots = lo g_{e} (1 + t)

, so grouping the powers of

a

and of

b

separately turns the given series into

lo g_{e} (1 + a) + lo g_{e} (1 + b) = lo g_{e} [(1 + a) (1 + b)]

From

tan^{- 1} a + tan^{- 1} b = \frac{π}{4}

with

0 < a, b < 1

\frac{a + b}{1 - ab} = 1

, so

a + b + ab = 1

and

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

The value of the series is

lo g_{e} 2

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