
If $y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} $, then $\dfrac{{dy}}{{dx}} = $?
A. $ - \dfrac{1}{2}$
B. $\dfrac{1}{2}$
C. $\dfrac{1}{{1 + {x^2}}}$
D. None of these
Answer
561.6k+ views
Hint: First transform the numerator using the formula $1 - \cos 2A = 2{\sin ^2}A$. After that transform the denominator using the formula $1 + \cos 2A = 2{\cos ^2}A$. Then do simplification, it will be in the form of $\dfrac{{\sin }}{{\cos }}$. Transform it in tan form and cancel out tan with ${\tan ^{ - 1}}$. Then differentiate it with respect to $x$ to get the desired result.
Complete step-by-step answer:
The given equation is $y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} $.
As we know,
$1 - \cos 2A = 2{\sin ^2}A$
And,
$1 + \cos 2A = 2{\cos ^2}A$
Use both formulae in the numerator and denominator of the square root,
$ \Rightarrow y = {\tan ^{ - 1}}\sqrt {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} $
Cancel out the common factor,
$ \Rightarrow y = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\sin }^2}\dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}}} $
As we know,
$\dfrac{{\sin A}}{{\cos A}} = \tan A$
Using this identity in the above equation, we will get
$ \Rightarrow y = {\tan ^{ - 1}}\sqrt {{{\tan }^2}\dfrac{x}{2}} $
Cancel out the square with square root,
$ \Rightarrow y = {\tan ^{ - 1}}\tan \dfrac{x}{2}$
Now cancel out tan with ${\tan ^{ - 1}}$, we will get
$ \Rightarrow y = \dfrac{x}{2}$
Now differentiate y with respect to x,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right)$
As we know,
$\dfrac{d}{{dx}}\left( {ax} \right) = a\dfrac{{dx}}{{dx}}$
Using the identity, we will get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{{dx}}{{dx}}$
Cancel out the common factor,
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{1}{2}$
Hence, the value of $\dfrac{{dy}}{{dx}}$ is $\dfrac{1}{2}$.
Note:
Trigonometry is an important branch of Mathematics. It mainly deals with triangles and their angles. It provides the relationships between the lengths and angles of triangles. It is the study of the relationships which involve angles, lengths, and heights of triangles.
Trigonometric formulas involve many trigonometric functions. These formulas and identities are true for all possible values of the variables. Trigonometric Ratios are also very basic to provide the relationship between the measurement of the angles and the length of the side of the right-angled triangle.
The six ratios which are the core of trigonometry are Sine (sin), Cosine (cos), Tangent (tan), Secant (sec), Cosecant (cosec), and Cotangent (cot)
The list of formulas for trigonometry are:-
Basic Formulas
Reciprocal Identities
Trigonometry Table
Periodic Identities
Co-function Identities
Sum and Difference Identities
Double Angle Identities
Triple Angle Identities
Half Angle Identities
Product Identities
Sum to Product Identities
Inverse Trigonometry Formulas
Complete step-by-step answer:
The given equation is $y = {\tan ^{ - 1}}\sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} $.
As we know,
$1 - \cos 2A = 2{\sin ^2}A$
And,
$1 + \cos 2A = 2{\cos ^2}A$
Use both formulae in the numerator and denominator of the square root,
$ \Rightarrow y = {\tan ^{ - 1}}\sqrt {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} $
Cancel out the common factor,
$ \Rightarrow y = {\tan ^{ - 1}}\sqrt {\dfrac{{{{\sin }^2}\dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2}}}} $
As we know,
$\dfrac{{\sin A}}{{\cos A}} = \tan A$
Using this identity in the above equation, we will get
$ \Rightarrow y = {\tan ^{ - 1}}\sqrt {{{\tan }^2}\dfrac{x}{2}} $
Cancel out the square with square root,
$ \Rightarrow y = {\tan ^{ - 1}}\tan \dfrac{x}{2}$
Now cancel out tan with ${\tan ^{ - 1}}$, we will get
$ \Rightarrow y = \dfrac{x}{2}$
Now differentiate y with respect to x,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{x}{2}} \right)$
As we know,
$\dfrac{d}{{dx}}\left( {ax} \right) = a\dfrac{{dx}}{{dx}}$
Using the identity, we will get,
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{2}\dfrac{{dx}}{{dx}}$
Cancel out the common factor,
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{1}{2}$
Hence, the value of $\dfrac{{dy}}{{dx}}$ is $\dfrac{1}{2}$.
Note:
Trigonometry is an important branch of Mathematics. It mainly deals with triangles and their angles. It provides the relationships between the lengths and angles of triangles. It is the study of the relationships which involve angles, lengths, and heights of triangles.
Trigonometric formulas involve many trigonometric functions. These formulas and identities are true for all possible values of the variables. Trigonometric Ratios are also very basic to provide the relationship between the measurement of the angles and the length of the side of the right-angled triangle.
The six ratios which are the core of trigonometry are Sine (sin), Cosine (cos), Tangent (tan), Secant (sec), Cosecant (cosec), and Cotangent (cot)
The list of formulas for trigonometry are:-
Basic Formulas
Reciprocal Identities
Trigonometry Table
Periodic Identities
Co-function Identities
Sum and Difference Identities
Double Angle Identities
Triple Angle Identities
Half Angle Identities
Product Identities
Sum to Product Identities
Inverse Trigonometry Formulas
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