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(A) Given $	an^{-1} a + 	an^{-1} b = \frac{\pi}{4}$ where $0 < a, b < 1$.Using the formula $	an^{-1} a + 	an^{-1} b = 	an^{-1} \left(\frac{a+b}{1

Vedclass · Accepted Answer

$\log_e 2$

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(A) Given $	an^{-1} a + 	an^{-1} b = \frac{\pi}{4}$ where $0 Using the formula $	an^{-1} a + 	an^{-1} b = 	an^{-1} \left(\frac{a+b}{1-ab}ight) = \frac{\pi}{4}$.Taking $	an$ on both sides,we get $\frac{a+b}{1-ab} = 	an \frac{\pi}{4} = 1$.Thus,$a+b = 1-ab$,which implies $a+b+ab = 1$.Adding $1$ to both sides,we get $1+a+b+ab = 2$,which factors as $(1+a)(1+b) = 2$.The given series is $S = \left(a - \frac{a^2}{2} + \frac{a^3}{3} - \dotsight) + \left(b - \frac{b^2}{2} + \frac{b^3}{3} - \dotsight)$.Using the logarithmic expansion $\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$ for $|x| $S = \log_e(1+a) + \log_e(1+b) = \log_e((1+a)(1+b))$.Substituting $(1+a)(1+b) = 2$,we get $S = \log_e 2$.

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

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