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Differentiate the following w.r.t x: $\tan^{-1}\bigg(\large {\frac {a \cos x - b \sin x}{b \cos x + a \sin x}}\bigg)$, where $\frac{-\pi}{2} < x < \frac{\pi}{2}$ and $\frac{a}{b} \tan x > -1$

$\begin{array}{1 1} 1 \\ \large\frac{1}{2} \\ -1 \\ \large\frac{-1}{2} \end{array} $

1 Answer

  • $\tan(A-B)=\large\frac{\tan A-\tan B}{1-\tan A\tan B}$
  • $\tan^{-1}(\tan x)=x$
  • $\large\frac{1-\tan x}{1+\tan x}$$=\tan(\large\frac{\pi}{4}\normalsize-4)$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
$y=\tan^{-1}\bigg[\large\frac{a\cos x-b\sin x}{b\cos x+a\sin x}\bigg]$
Consider $\large\frac{a\cos x-b\sin x}{b\cos x+a\sin x}$
Divide throughout by $b\cos x$
$\Rightarrow \large\frac{\Large\frac{a}{b}-\tan x}{1+\Large\frac{a\tan x}{b}}$
This is of the form $\tan(\large\frac{\pi}{4}\frac{a}{b}-x)$
Step 2:
$y=\tan^{-1}\bigg(\tan(\large\frac{\pi}{4}\frac{a}{b}-\normalsize x)\bigg)$
Differentiating w.r.t $x$ we get,
answered Jul 1, 2013 by sreemathi.v

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