Question

please solve this (specifically for 333333)​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow 1$

Step-by-step explanation:

Given :

$\displaystyle \sf \frac{d}{dx}\left\{\tan^{-1}\left(\frac{a\sin x+b\cos x}{a\cos x-b \sin x}\right)\right\}$

We have to find it derivative :

Dividing by a cos x we get :

$\displaystyle \sf \longrightarrow \frac{d}{dx}\left\{\tan^{-1}\left(\frac{\dfrac{a\sin x}{a\cos x} +\dfrac{b\cos x}{a\cos x}}{\dfrac{a\cos x}{a\cos x}- \dfrac{b\sin x}{a\cos x}}\right)\right\}$

$\displaystyle \sf \longrightarrow \frac{d}{dx}\left\{\tan^{-1}\left(\frac{\tan x +\dfrac{b}{a}}{ 1 -\dfrac{b}{a}.\tan x}\right)\right\}$

We know :

$\displaystyle \sf \tan^{-1} A+\tan^{-1} B=\tan^{-1}\left(\frac{A+B}{1-A.B} \right)$

$\displaystyle \sf \longrightarrow \frac{d}{dx}\left\{\tan^{-1}\left(\tan x \right)+\tan^{-1}\left( \frac{b}{a} \right)\right\}$

We also know :

$\displaystyle \sf \tan^{-1} (\tan \theta) = \theta$

$\displaystyle \sf \longrightarrow \frac{d}{dx}\left\{x+\tan^{-1}\left( \frac{b}{a} \right)\right\}$

Since there is no x variable in tan inverse term :

It will consider as constant. We know derivative of constant is zero.

$\displaystyle \sf \longrightarrow \frac{d}{dx} (x)+\frac{d}{dx}\left(\tan^{-1}\left( \frac{b}{a} \right)\right)$

$\displaystyle \sf \longrightarrow 1+0$

$\displaystyle \sf \longrightarrow 1$

Hence we get required answer.

Answer 2

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