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

$\frac{d}{d x}\left(\cos ^{-1}\left(\frac{x-\frac{1}{x}}{x+\frac{1}{x}}\right)\right)=$

$\frac{d}{{dy}}\left( {{{\sin }^{ - 1}}\left( {\frac{{3y}}{2} - \frac{{{y^3}}}{2}} \right)} \right) = $

The differential coefficient of ${\tan ^{ - 1}}\left( {\frac{{\sqrt {1 + {x^2}} - 1}}{x}} \right)$ with respect to ${\tan ^{ - 1}}x$ is

Find the derivative: $\frac{d}{dx} \tan^{-1}(\sec x + \tan x) = $

Find the derivative of $\cos^{-1}\sqrt{\cos x}$ with respect to $x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module