Question

If \[y = {\tan ^{ - 1}}\sqrt {\dfrac{{\left( {1 - \sin x} \right)}}{{\left( {1 + \sin x} \right)}}} \] then the value of \[\dfrac{{dy}}{{dx}}\] at \[x = \dfrac{\pi }{6}\] is
(1) \[\dfrac{1}{2}\]
(2) \[ - \dfrac{1}{2}\]
(3) \[1\]
(4) \[ - 1\]

Answer 1

Hint: We have to differentiate \[y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \sin x}}{{1 + \sin x}}} \] with respect to \[x\] to find the value of \[\dfrac{{dy}}{{dx}}\] and then we have to put the value of \[x = \dfrac{\pi }{6}\] to get the final result. First, we will simplify this using the periodicity identity i.e., \[\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\] and we will use the double angle identities i.e., \[\cos 2\theta = 1 - 2{\sin ^2}\theta \] and \[\cos 2\theta = 2{\cos ^2}\theta - 1\] then we will use the formula of \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] and finally we will use inverse trigonometric formula i.e., \[{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta \] then we differentiate the term obtained after simplification and put the given value of \[x\]

Complete answer:Let,
It is given that, \[y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \sin x}}{{1 + \sin x}}} {\text{ }} - - - \left( 1 \right)\]
Now, use the periodicity identity
i.e., \[\sin x = \cos \left( {\dfrac{\pi }{2} - x} \right)\]
\[ \Rightarrow 1 - \sin x = 1 - \cos \left( {\dfrac{\pi }{2} - x} \right)\] and
 \[1 + \sin x = 1 + \cos \left( {\dfrac{\pi }{2} - x} \right)\]
\[\therefore \] equation \[\left( 1 \right)\] becomes,
\[y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos \left( {\dfrac{\pi }{2} - x} \right)}}{{1 + \cos \left( {\dfrac{\pi }{2} - x} \right)}}} {\text{ }} - - - \left( 2 \right)\]
Now, use the double angle identities,
i.e., \[\cos 2\theta = 1 - 2{\sin ^2}\theta \]
\[ \Rightarrow 1 - \cos 2\theta = 2{\sin ^2}\theta {\text{ }} - - - \left( a \right)\]
And \[\cos 2\theta = 2{\cos ^2}\theta - 1\]
\[ \Rightarrow 1 + \cos 2\theta = 2{\cos ^2}\theta {\text{ }} - - - \left( b \right)\]
Now, from equation \[\left( 2 \right)\]
\[2\theta = \left( {\dfrac{\pi }{2} - x} \right)\]
\[ \Rightarrow \theta = \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
\[\therefore \] from \[\left( a \right)\] when we put the values, we get
\[1 - \cos \left( {\dfrac{\pi }{2} - x} \right) = 2{\sin ^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
and from \[\left( b \right)\] we get,
\[1 + \cos \left( {\dfrac{\pi }{2} - x} \right) = 2{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
So, equation \[\left( 2 \right)\] becomes
\[y = {\tan ^{ - 1}}\sqrt {\dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}} \]
\[ \Rightarrow y = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}} {\text{ }} - - - \left( 3 \right)\]
We know that \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[ \Rightarrow {\tan ^2}\theta = \dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}\]
Here, \[\theta = \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
\[ \Rightarrow {\tan ^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}{{{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)}}\]
\[\therefore \] equation \[\left( 3 \right)\] becomes
\[y = {\tan ^{ - 1}}\sqrt {{{\tan }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} {\text{ }} - - - \left( 4 \right)\]
Now, we know that
\[\sqrt {{x^2}} = x\]
\[\therefore \sqrt {{{\tan }^2}\theta } = \tan \theta \]
Here, \[\theta = \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
\[ \Rightarrow \sqrt {{{\tan }^2}\left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} = \tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
\[\therefore \] equation \[\left( 4 \right)\] becomes,
\[y = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right){\text{ }} - - - \left( 5 \right)\]
Now use inverse trigonometric formula, \[{\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta \]
\[ \Rightarrow {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right) = \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
\[\therefore \] equation \[\left( 5 \right)\] becomes,
\[y = \dfrac{\pi }{4} - \dfrac{x}{2}{\text{ }} - - - \left( 6 \right)\]
We know that \[\dfrac{{d(c)}}{{dx}} = 0\] where c is constant
And \[\dfrac{{dx}}{{dx}} = 1\]
\[\therefore \] on differentiating equation \[\left( 6 \right)\] we get,
\[\dfrac{{dy}}{{dx}} = 0 - \dfrac{1}{2}\]
\[\because \dfrac{\pi }{4}\] is a constant term
\[ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{2}\]
As there are no terms of \[x\] in \[\dfrac{{dy}}{{dx}}\]
\[\therefore \] at \[x = \dfrac{\pi }{6}\] also \[\dfrac{{dy}}{{dx}} = - \dfrac{1}{2}\]
Hence, option \[\left( 2 \right)\] is correct.

Note:
When we are faced with trigonometric questions, think about each identity and properties. Start by identifying which one we really need to use to solve the question. Then, we can quickly determine the best way to simplify the problem and can find a solution.
There is an alternative way to solve this question:
We have given, \[y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \sin x}}{{1 + \sin x}}} {\text{ }} - - - \left( 1 \right)\]
We know that \[{\cos ^2}\theta + {\sin ^2}\theta = 1\] which can also be written in the form of half angle as
\[{\cos ^2}\dfrac{\theta }{2} + {\sin ^2}\dfrac{\theta }{2} = 1{\text{ }} - - - \left( a \right)\]
and \[\sin 2\theta = 2\sin \theta \cos \theta \] which can also be written in the form of half angle as
\[\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}{\text{ }} - - - \left( b \right)\]
Now, substitute the values of \[\left( a \right)\] and \[\left( b \right)\] in equation \[\left( 1 \right)\] we get
\[y = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\cos }^2}\dfrac{x}{2} + {{\sin }^2}\dfrac{x}{2} - 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2} + {{\sin }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}} {\text{ }} - - - \left( 2 \right)\]
We know that,
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
So, equation \[\left( 2 \right)\] becomes,
\[y = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}^2}}}{{{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}^2}}}} \]
Cancelling square root, we get
\[y = {\tan ^{ - 1}}\left( {\dfrac{{\cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}}} \right)\]
Dividing by \[\cos \dfrac{x}{2}\] in numerator and denominator both, we get
\[y = {\tan ^{ - 1}}\left( {\dfrac{{1 - \tan \dfrac{x}{2}}}{{1 + \tan \dfrac{x}{2}}}} \right){\text{ }} - - - \left( 3 \right)\]
We know that \[\tan \left( {\dfrac{\pi }{4} - x} \right) = \dfrac{{1 - \tan x}}{{1 + \tan x}}\]
So, equation \[\left( 3 \right)\] becomes,
\[y = {\tan ^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)} \right)\]
\[ \Rightarrow y = \left( {\dfrac{\pi }{4} - \dfrac{x}{2}} \right)\]
On differentiating,
\[\dfrac{{dy}}{{dx}} = - \dfrac{1}{2}\]
As there are no terms of \[x\] in \[\dfrac{{dy}}{{dx}}\]
\[\therefore \] at \[x = \dfrac{\pi }{6}\] also \[\dfrac{{dy}}{{dx}} = - \dfrac{1}{2}\]
Hence, option \[\left( 2 \right)\] is correct.