Question:
If $ y={{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right), $ then $ \frac{dy}{dx} $ is equal to
If $ y={{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right), $ then $ \frac{dy}{dx} $ is equal to
Updated On: Jun 8, 2024
- $ \frac{1}{2} $
- $ 2 $
- $ -2 $
- $ -\frac{1}{2} $
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The Correct Option is D
Solution and Explanation
Given that, $ y={{\tan }^{-1}}\left( \frac{\cos x}{1+\sin x} \right) $
$={{\tan }^{-1}}\left( \frac{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}} \right) $
$={{\tan }^{-1}}\left( \frac{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}{{{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}} \right) $
$={{\tan }^{-1}}\left( \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} \right) $
$={{\tan }^{-1}}\left( \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \right) $
$={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{4}-\frac{x}{2} \right) \right) $
$=\frac{\pi }{4}-\frac{x}{2} $
$ \therefore $ $ y=\frac{\pi }{4}-\frac{x}{2} $
Differentiating it w.r.t., $ x $ we get $ \frac{dy}{dx}=-\frac{1}{2} $
$={{\tan }^{-1}}\left( \frac{{{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2}}{{{\sin }^{2}}\frac{x}{2}+{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}.\cos \frac{x}{2}} \right) $
$={{\tan }^{-1}}\left( \frac{\left( \cos \frac{x}{2}-\sin \frac{x}{2} \right)\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}{{{\left( \cos \frac{x}{2}+\sin \frac{x}{2} \right)}^{2}}} \right) $
$={{\tan }^{-1}}\left( \frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} \right) $
$={{\tan }^{-1}}\left( \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}} \right) $
$={{\tan }^{-1}}\left( \tan \left( \frac{\pi }{4}-\frac{x}{2} \right) \right) $
$=\frac{\pi }{4}-\frac{x}{2} $
$ \therefore $ $ y=\frac{\pi }{4}-\frac{x}{2} $
Differentiating it w.r.t., $ x $ we get $ \frac{dy}{dx}=-\frac{1}{2} $
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.