
Differentiate $ {\tan ^{ - 1}}\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right) $ with respect to $ x $ .
Answer
539.7k+ views
Hint: se trigonometric expansion accordingly and solve and then differentiate.
First, we are going to write the denominator, as we know a formula of $ 1 + \cos x = 2\dfrac{{{{\cos }^2}x}}{2} $ and also for the convenience of the numerator, we will convert denominator in the same form using the formula of $ \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} $ and then we will write these into the given expression and simplify if after simplifying we will use inverse trigonometric properties and get the value and then we apply differentiation to the expression with respect to x.
Complete step-by-step answer:
We are given an expression $ {\tan ^{ - 1}}\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right) $
Let’s consider it as $ y = {\tan ^{ - 1}}\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right) $
Now, let us consider the numerator which is $ 1 + \cos x $ , we have a expansion which can be used in it place which is $ 1 + \cos x = 2\dfrac{{{{\cos }^2}x}}{2} $ .
Now, coming to the denominator which is $ \sin x $ and for the convenience of the numerator, we can convert this into the same form with the expansion of $ \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} $ .
Now, we are going to substitute them in the actual expression.
$ y = {\tan ^{ - 1}}\left( {\dfrac{{2\dfrac{{{{\cos }^2}x}}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} \right) $
Now, we can cancel the like terms in numerator and denominator, then we get
$ y = {\tan ^{ - 1}}(\dfrac{{\cot x}}{2}) $
We can write the cot function in terms of tan by
$ \dfrac{{\cot x}}{2} = \tan \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right) $
Substitute in the above equation. We get
$ y = {\tan ^{ - 1}}(\tan \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right)) $
Here the inverse function and the normal function gets cancelled, leaving
$ y = \dfrac{\pi }{2} - \dfrac{x}{2} $
Now, we can easily differentiate it, we get
$ \dfrac{{dy}}{{dx}} = - \dfrac{1}{2} $
This the required value of the given expression.
So, the correct answer is “ $ - \dfrac{1}{2} $ ”.
Note: Questions of derivative with trigonometry concepts require pure skill in which you need to use your concepts and either assume/put values or use some mathematical operations on the given form to get another form which is somehow connected to the required answer. It can be a direct formula or an even simplified version, you must convert it to a simple form using properties.
First, we are going to write the denominator, as we know a formula of $ 1 + \cos x = 2\dfrac{{{{\cos }^2}x}}{2} $ and also for the convenience of the numerator, we will convert denominator in the same form using the formula of $ \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} $ and then we will write these into the given expression and simplify if after simplifying we will use inverse trigonometric properties and get the value and then we apply differentiation to the expression with respect to x.
Complete step-by-step answer:
We are given an expression $ {\tan ^{ - 1}}\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right) $
Let’s consider it as $ y = {\tan ^{ - 1}}\left( {\dfrac{{1 + \cos x}}{{\sin x}}} \right) $
Now, let us consider the numerator which is $ 1 + \cos x $ , we have a expansion which can be used in it place which is $ 1 + \cos x = 2\dfrac{{{{\cos }^2}x}}{2} $ .
Now, coming to the denominator which is $ \sin x $ and for the convenience of the numerator, we can convert this into the same form with the expansion of $ \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2} $ .
Now, we are going to substitute them in the actual expression.
$ y = {\tan ^{ - 1}}\left( {\dfrac{{2\dfrac{{{{\cos }^2}x}}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} \right) $
Now, we can cancel the like terms in numerator and denominator, then we get
$ y = {\tan ^{ - 1}}(\dfrac{{\cot x}}{2}) $
We can write the cot function in terms of tan by
$ \dfrac{{\cot x}}{2} = \tan \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right) $
Substitute in the above equation. We get
$ y = {\tan ^{ - 1}}(\tan \left( {\dfrac{\pi }{2} - \dfrac{x}{2}} \right)) $
Here the inverse function and the normal function gets cancelled, leaving
$ y = \dfrac{\pi }{2} - \dfrac{x}{2} $
Now, we can easily differentiate it, we get
$ \dfrac{{dy}}{{dx}} = - \dfrac{1}{2} $
This the required value of the given expression.
So, the correct answer is “ $ - \dfrac{1}{2} $ ”.
Note: Questions of derivative with trigonometry concepts require pure skill in which you need to use your concepts and either assume/put values or use some mathematical operations on the given form to get another form which is somehow connected to the required answer. It can be a direct formula or an even simplified version, you must convert it to a simple form using properties.
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