#### Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.11 Question 4 Maths Textbook Solution.

$\tan^{-1}2^{n}-\frac{\pi }{4}$
Hint:
We have to focus on calculation of infinite series.
Given:\begin{aligned} &\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{2}{9}\right)+\ldots+\left.\tan ^{-1}\left(\frac{2^{n-1}}{1+2^{2 n-1}}\right)\right|_{-1} \\ \end{aligned}

Solution:
\begin{aligned} &\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{2}{9}\right)+\ldots+\left.\tan ^{-1}\left(\frac{2^{n-1}}{1+2^{2 n-1}}\right)\right|_{-1} \\ &=\tan ^{-1}\left(\frac{2-1}{1+2 x 1}\right)+\tan ^{-1}\left(\frac{4-2}{1+4 x 2}\right)+\ldots+\tan ^{-1}\left(\frac{2^{n}-2^{n-1}}{1+2^{n} \cdot 2^{n-1}}\right) \\ &=\left(\tan ^{-1} 2-\tan ^{-1} 1\right)+\left(\tan ^{-1} 4-\tan ^{-1} 2\right)+\ldots .+\left(\tan ^{-1} 2^{n}-\tan ^{-1} 2^{n-1}\right) \\ &=\tan ^{-1} 2^{n}-\tan ^{-1} 1 \\ &=\tan ^{-1} 2^{n}-\frac{\pi}{4} \end{aligned}

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