KCET · Maths · Inverse Trigonometric Functions
\(\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots\) upto \(\infty\) is equal to
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{3}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\pi}{5}\)
Answer & Solution
Correct Answer
(A) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots \infty\)
\(=\sum_{\mathrm{r}=1}^{\infty} \cot ^{-1}\left(2 \cdot \mathrm{r}^{2}\right)=\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 \mathrm{r}^{2}}\right)\)
\(=\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1}\left(\frac{(1+2 \mathrm{r})+(1-2 \mathrm{r})}{1-(1+2 \mathrm{r})(1-2 \mathrm{r})}\right)\)
\(=\sum_{\mathrm{r}=1}^{\infty}\left[\tan ^{-1}(1+2 \mathrm{r})+\tan ^{-1}(1-2 \mathrm{r})\right]\)
\(=\tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 5-\tan ^{-1} 3+\tan ^{-1} 7-\tan ^{-1} 5+\ldots+\tan ^{-1} \infty\)
\(=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}\)
\(=\sum_{\mathrm{r}=1}^{\infty} \cot ^{-1}\left(2 \cdot \mathrm{r}^{2}\right)=\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 \mathrm{r}^{2}}\right)\)
\(=\sum_{\mathrm{r}=1}^{\infty} \tan ^{-1}\left(\frac{(1+2 \mathrm{r})+(1-2 \mathrm{r})}{1-(1+2 \mathrm{r})(1-2 \mathrm{r})}\right)\)
\(=\sum_{\mathrm{r}=1}^{\infty}\left[\tan ^{-1}(1+2 \mathrm{r})+\tan ^{-1}(1-2 \mathrm{r})\right]\)
\(=\tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 5-\tan ^{-1} 3+\tan ^{-1} 7-\tan ^{-1} 5+\ldots+\tan ^{-1} \infty\)
\(=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}\)
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