Question

What is the value of \[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\]?
A.\[cosec\,x\]
B.\[secx\]
C.\[\cos ec\dfrac{x}{2}\]
D.\[sec\dfrac{x}{2}\]

Answer 1

Hint: To solve the question we will first apply the formula \[1 - \cos x = 2{\sin ^2}x\] in numerator and \[1 + \cos x = 2{\cos ^2}x\] in denominator. Then simplify the denominator and numerator. Then apply chain on \[\tan \dfrac{x}{2}\] to get the desire result.

Formula used:
We will use the trigonometric formulas \[1 - \cos x = 2{\sin ^2}\dfrac{x}{2}\], \[1 + \cos x = 2{\cos ^2}\dfrac{x}{2}\], \[\tan x = \dfrac{{\sin x}}{{\cos x}}\]and \[\sec x = \dfrac{1}{{\cos x}}\], \[\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\],\[\dfrac{1}{{\sin x}} = cosec\,x\] and the derivative formulas \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\], \[\dfrac{d}{{dx}}\tan x = {\sec ^2}\dfrac{x}{2}\] and \[\dfrac{d}{{dx}}x = 1\], and we will use the chain rule.

Complete Step-by-Step Solution:
Given \[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right]\]
Now we will use the trigonometric formulas, \[1 - \cos x = 2{\sin ^2}\dfrac{x}{2}\], \[1 + \cos x = 2{\cos ^2}\dfrac{x}{2}\], and substituting the values in the given expression,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{2{{\sin }^2}\dfrac{x}{2}}}{{2{{\cos }^2}\dfrac{x}{2}}}} } \right]\]
Now use know that\[\tan x = \dfrac{{\sin x}}{{\cos x}}\], now we will simplify using the formula,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{d}{{dx}}\left[ {\log \sqrt {{{\tan }^2}\dfrac{x}{2}} } \right]\]
Now we will take the square root, we will get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{d}{{dx}}\left[ {\log \left( {\tan \dfrac{x}{2}} \right)} \right]\]
Now using the derivative formula \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\] and chain rule,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{{\tan \dfrac{x}{2}}}\dfrac{d}{{dx}}\tan \dfrac{x}{2}\]
Now again we will use derivative formula \[\dfrac{d}{{dx}}\tan x = {\sec ^2}\dfrac{x}{2}\], and we will use the chain rule,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{{\tan \dfrac{x}{2}}}\left( {{{\sec }^2}\dfrac{x}{2}} \right)\dfrac{d}{{dx}}\dfrac{x}{2}\]
Now again using the derivative formula, \[\dfrac{d}{{dx}}x = 1\], we will get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{{\tan \dfrac{x}{2}}}\left( {{{\sec }^2}\dfrac{x}{2}} \right)\left( {\dfrac{1}{2}} \right)\]\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{{\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}\left( {\dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}}} \right)\left( {\dfrac{1}{2}} \right)\]
Now we will use the trigonometric formulas and \[\sec x = \dfrac{1}{{\cos x}}\], we will get,

Now we will simplify the expression we will get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{{\cos \dfrac{x}{2}}}{{\sin \dfrac{x}{2}}}\left( {\dfrac{1}{{{{\cos }^2}\dfrac{x}{2}}}} \right)\left( {\dfrac{1}{2}} \right)\],
Now we will eliminate the like terms, we will get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{{\sin \dfrac{x}{2}}}\left( {\dfrac{1}{{\cos \dfrac{x}{2}}}} \right)\left( {\dfrac{1}{2}} \right)\]
Now we will simplify the expression we will get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}\],
Now we will use the trigonometric identity \[\sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\],we will get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{{\sin x}}\],
Now using the trigonometric formula, \[\dfrac{1}{{\sin x}} = cosec\,x\], we will get,
\[ \Rightarrow \dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = cosec\,x\],

The correct option is A.

Note: There is another method to solve the problem. That is shown below:
Apply the formula \[\log {a^m} = m\log a\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{d}{{dx}}\left[ {\dfrac{1}{2}\log \dfrac{{1 - \cos x}}{{1 + \cos x}}} \right]\]
Apply chain rule
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{2}\dfrac{1}{{\dfrac{{1 - \cos x}}{{1 + \cos x}}}}\dfrac{d}{{dx}}\left[ {\dfrac{{1 - \cos x}}{{1 + \cos x}}} \right]\]
Apply product rule \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{2} \cdot \dfrac{{1 + \cos x}}{{1 - \cos x}} \cdot \dfrac{{\left( {1 + \cos x} \right)\dfrac{d}{{dx}}\left( {1 - \cos x} \right) - \left( {1 - \cos x} \right)\dfrac{d}{{dx}}\left( {1 + \cos x} \right)}}{{{{\left( {1 + \cos x} \right)}^2}}}\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{2} \cdot \dfrac{{1 + \cos x}}{{1 - \cos x}} \cdot \dfrac{{\left( {1 + \cos x} \right)\left( {\sin x} \right) - \left( {1 - \cos x} \right)\left( { - \sin x} \right)}}{{{{\left( {1 + \cos x} \right)}^2}}}\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{1}{2} \cdot \dfrac{{1 + \cos x}}{{1 - \cos x}} \cdot \dfrac{{\sin x + \sin x\cos x + \sin x - \sin x\cos x}}{{{{\left( {1 + \cos x} \right)}^2}}}\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{{2\sin x}}{{2\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)}}\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{{\sin x}}{{1 - {{\cos }^2}x}}\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = \dfrac{{\sin x}}{{{{\sin }^2}x}}\]
\[\dfrac{d}{{dx}}\left[ {\log \sqrt {\dfrac{{1 - \cos x}}{{1 + \cos x}}} } \right] = cosec x\]