Trignometry - some inverse problems....

1

nkhlshd ·2012-04-27 04:31:04

use arctan(1/x^2+x+1) = arctan(x+1) - arctan(x)
arctan(1/x^2+3x+3)= arctan(x+2) - arctan(x+1)....and so on..summing up to n terms, terms will cancel out , leaving the final sum as
arctan(x+n)-arctan(x)
=arctan (n/x^2+nx+1)

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1

nkhlshd ·2012-04-27 04:35:25

x = -0.5

converting arccot to arcsin and arctan to arccos..we get
1/âˆš[1+(x+1)^2] =1/âˆš[1+x^2]

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man111 singh ·2012-04-27 04:45:07

\hspace{-16}\bf{(3)\;\;:: \sum_{n=1}^{10}\;\sum_{m=1}^{10}\tan^{-1}\left(\frac{m}{n}\right)=}$\\\\\\ $\bf{\sum_{n=1}^{10}\left\{\tan^{-1}\left(\frac{1}{n}\right)+ \tan^{-1}\left(\frac{2}{n}\right)+.......+\tan^{-1}\left(\frac{10}{n}\right)\right\}}$\\\\\\ $\bf{\sum_{n=1}^{10}\tan^{-1}\left(\frac{1}{n}\right)+\sum_{n=1}^{10}\tan^{-1}\left(\frac{2}{n}\right)+.......+\sum_{n=1}^{10}\tan^{-1}\left(\frac{10}{n}\right)}$\\\\\\

\hspace{-10}$\begin{Bmatrix} \bf{\tan^{-1}\left(\frac{1}{1}\right)} +\bf{\tan^{-1}\left(\frac{1}{2}\right)} +.......+\bf{\tan^{-1}\left(\frac{1}{10}\right)} \\\\ \bf{\tan^{-1}\left(\frac{2}{1}\right)+\tan^{-1}\left(\frac{2}{2}\right)+......+\tan^{-1}\left(\frac{2}{10}\right)}\\\\ \bf{\tan^{-1}\left(\frac{3}{1}\right)+\tan^{-1}\left(\frac{3}{2}\right)+......+\tan^{-1}\left(\frac{3}{10}\right)}\\\\ \bf{\tan^{-1}\left(\frac{4}{1}\right)+\tan^{-1}\left(\frac{4}{2}\right)+......+\tan^{-1}\left(\frac{4}{10}\right)}\\\\ .............................................................. \\ \\ .............................................................. \\\\ .............................................................. \\ \\ .............................................................. \\\\ .............................................................. \\\\ \bf{\tan^{-1}\left(\frac{10}{1}\right)+\tan^{-1}\left(\frac{10}{2}\right)+......+\tan^{-1}\left(\frac{10}{10}\right)}\\\\ \end{Bmatrix}$

[\hspace{-10}$So It is a Rectangular Arrangement of $\bf{10 \times 10}$ array\\\\\\ Here $\bf{10}$ values of $\bf{\tan^{-1}(1)=\tan^{-1}\left(tan \frac{\pi}{4}\right)=\frac{\pi}{4}}$\\\\\\ So $\bf{10 \times \frac{\pi}{4}=\frac{10\pi}{4}=\frac{5\pi}{2}}$\\\\\\ And Remaining $\bf{90}$ values are in the form of\\\\\\ $\bf{\tan^{-1}(a)}$ and $\bf{\tan^{-1}\left(\frac{1}{a}\right)}$\\\\\\ Where $\bf{a>0}$\\\\ So $\bf{\tan^{-1}(a)+\tan^{-1}\left(\frac{1}{a}\right)=\frac{\pi}{2}}$\\\\\\ So Total These Type of $\bf{45}$ Pair\\\\\\ So $\bf{45 \times \frac{\pi}{2}=\frac{45\pi}{2}}$\\\\\\ So Total Sum of Elements of $\bf{10 \times 10}$ array is \\\\\\ $\bf{\frac{5\pi}{2}+\frac{45\pi}{2}=\frac{50\pi}{2}=25\pi}$\\\\\\ So $\boxed{\boxed{\bf{\bf{k=25}}}}$