# If $0, and $\tan ^{-1}a + \tan ^{-1}b =\frac{\pi}{4}$, then the value of  $(a+b)-\left ( \frac{a^2+b^2}{2} \right )+\left ( \frac{a^3+b^3}{3} \right )-\left ( \frac{a^4+b^4}{4} \right )+.........$ is: Option: 1 e Option: 3 $\log _e\left (\frac{e}{2} \right )$ Option: 5 $\log _e 2$ Option: 7 $e^2-1$

$\\\tan ^{-1} \mathrm{a}+\tan ^{-1} \mathrm{~b}=\frac{\pi}{4} \quad\quad 0<\mathrm{a}, \mathrm{b}<1 \\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}=1 \\ \mathrm{a}+\mathrm{b}=1-\mathrm{ab} \\ (\mathrm{a}+1)(\mathrm{b}+1)=2$

$\\\text { Now }\left[a-\frac{a^{2}}{2}+\frac{a^{3}}{3}+\ldots\right]+\left[b-\frac{b^{2}}{2}+\frac{b^{3}}{3}+\ldots\right] \\ =\log _{e}(1+a)+\log _{e}(1+b) \\ \left(\because \text { expansion of } \log _{e}(1+x)\right) \\ =\log _{e}[(1+a)(1+b)] \\ =\log _{e} 2$

## Similar Questions

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
##### Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
##### Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-
The text passage contains a mathematical question involving a series and the inverse tangent function, followed by a detailed solution and a set of answer options. The solution is provided step-by-step, leading to the final answer.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# If $0, and $\tan ^{-1}a + \tan ^{-1}b =\frac{\pi}{4}$, then the value of  $(a+b)-\left ( \frac{a^2+b^2}{2} \right )+\left ( \frac{a^3+b^3}{3} \right )-\left ( \frac{a^4+b^4}{4} \right )+.........$ is: Option: 1 e Option: 3 $\log _e\left (\frac{e}{2} \right )$ Option: 5 $\log _e 2$ Option: 7 $e^2-1$

$\\\tan ^{-1} \mathrm{a}+\tan ^{-1} \mathrm{~b}=\frac{\pi}{4} \quad\quad 0<\mathrm{a}, \mathrm{b}<1 \\ \Rightarrow \frac{\mathrm{a}+\mathrm{b}}{1-\mathrm{ab}}=1 \\ \mathrm{a}+\mathrm{b}=1-\mathrm{ab} \\ (\mathrm{a}+1)(\mathrm{b}+1)=2$

$\\\text { Now }\left[a-\frac{a^{2}}{2}+\frac{a^{3}}{3}+\ldots\right]+\left[b-\frac{b^{2}}{2}+\frac{b^{3}}{3}+\ldots\right] \\ =\log _{e}(1+a)+\log _{e}(1+b) \\ \left(\because \text { expansion of } \log _{e}(1+x)\right) \\ =\log _{e}[(1+a)(1+b)] \\ =\log _{e} 2$

## Similar Questions

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
##### Test Series JEE Main May 2021

Unlimited Chapter Wise Tests, Unlimited Subject Wise Tests, Unlimited Full Mock Tests, Get Personalized Performance Analysis Report,.

₹ 6999/- ₹ 2999/-
##### Knockout JEE Main May 2022

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 34999/- ₹ 14999/-
user
The text passage contains a mathematical question involving a series and the inverse tangent function, followed by a detailed solution and a set of answer options. The solution is provided step-by-step, leading to the final answer.

TYPE: SINGLE_QUESTION_WITH_ANSWER