If f(x) = tan^{-1}left(frac{1}{sin^2 x + sin | vedclass

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(C) The general term of the series is $T_n = 	an^{-1}\left(\frac{1}{\sin^2 x + (2n-1)\sin x + (n^2-n+1)}ight)$.We can rewrite this as $T_n = 	an^{

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$\frac{-100}{101}$

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(C) The general term of the series is $T_n = 	an^{-1}\left(\frac{1}{\sin^2 x + (2n-1)\sin x + (n^2-n+1)}ight)$.We can rewrite this as $T_n = 	an^{-1}\left(\frac{(\sin x + n) - (\sin x + n - 1)}{1 + (\sin x + n)(\sin x + n - 1)}ight)$.Using the identity $	an^{-1} A - 	an^{-1} B = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$, we get $T_n = 	an^{-1}(\sin x + n) - 	an^{-1}(\sin x + n - 1)$.Summing up to $10$ terms, we have $f(x) = \sum_{n=1}^{10} [	an^{-1}(\sin x + n) - 	an^{-1}(\sin x + n - 1)]$.This is a telescoping series, so $f(x) = 	an^{-1}(\sin x + 10) - 	an^{-1}(\sin x)$.Differentiating with respect to $x$, we get $f'(x) = \frac{\cos x}{1 + (\sin x + 10)^2} - \frac{\cos x}{1 + \sin^2 x}$.At $x = 0$, $f'(0) = \frac{\cos 0}{1 + (0 + 10)^2} - \frac{\cos 0}{1 + 0^2} = \frac{1}{101} - 1 = \frac{-100}{101}$.

If $f(x) = \tan^{-1}\left(\frac{1}{\sin^2 x + \sin x + 1}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 3\sin x + 3}\right) + \tan^{-1}\left(\frac{1}{\sin^2 x + 5\sin x + 7}\right) + \dots$ up to $10$ terms, then $f'(0) = $

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