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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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^{

Vedclass · Accepted Answer

$\frac{-100}{101}$

Vedclass · Answer

(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) = $

Similar Questions

The derivative of $\operatorname{Sec}^{-1}\left(\frac{1}{2x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is

If $y=\tan ^{-1}\left(\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right)$,then $\left(\frac{d y}{d x}\right)$ at $x=0$ is

If $y = \sin^{-1}(\sqrt{x})$,then $\frac{dy}{dx} = $

If $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left[ \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} \right]$,then $\frac{dy}{dx} = $

If $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module