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If $y=\tan ^{-1}\left[\frac{5 \cos x-12 \sin x}{12 \cos x+5 \sin x}\right]$, then $\frac{d y}{d x}$ is equal to
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Verified Answer
The correct answer is:
-1
Given,
$$
\begin{aligned}
y & =\tan ^{-1}\left[\frac{5 \cos x-12 \sin x}{12 \cos x+5 \sin x}\right] \\
& =\tan ^{-1}\left[\frac{\frac{5}{12}-\tan x}{1+\frac{5}{12} \tan x}\right]
\end{aligned}
$$
[divide by $12 \cos x$ in denominator and numerator]
$$
\begin{aligned}
& =\tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(\tan x) \\
y & =\tan ^{-1}\left(\frac{5}{12}\right)-x
\end{aligned}
$$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left\{\tan ^{-1}\left(\frac{5}{12}\right)\right\}-\frac{d}{d x}(x)=0-1 \\
& \therefore \quad \frac{d y}{d x}=-1
\end{aligned}
$$
$$
\begin{aligned}
y & =\tan ^{-1}\left[\frac{5 \cos x-12 \sin x}{12 \cos x+5 \sin x}\right] \\
& =\tan ^{-1}\left[\frac{\frac{5}{12}-\tan x}{1+\frac{5}{12} \tan x}\right]
\end{aligned}
$$
[divide by $12 \cos x$ in denominator and numerator]
$$
\begin{aligned}
& =\tan ^{-1}\left(\frac{5}{12}\right)-\tan ^{-1}(\tan x) \\
y & =\tan ^{-1}\left(\frac{5}{12}\right)-x
\end{aligned}
$$
On differentiating both sides w.r.t. $x$, we get
$$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left\{\tan ^{-1}\left(\frac{5}{12}\right)\right\}-\frac{d}{d x}(x)=0-1 \\
& \therefore \quad \frac{d y}{d x}=-1
\end{aligned}
$$
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