
How do you prove $\ln (\sec \theta ) = - \ln (\cos \theta )$?
Answer
532.2k+ views
Hint: In the above question, we need to prove that the right hand side is equal to the left hand side.
So firstly consider the left hand side $\ln (\sec \theta )$. We know that secant is the inverse function of cosine function. So write $\sec \theta $ in terms of $\cos \theta $. Then we make use of properties of logarithm. In this we use division property given by $\ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b$. Substitute for the values of a and b to simplify and obtain the result which must be equal to the right hand side.
Complete step by step solution:
Given an equation $\ln (\sec \theta ) = - \ln (\cos \theta )$ …… (1)
We are asked to prove the left hand side is equal to the right hand side.
So let us begin with the L.H.S.
Consider the L.H.S. given by $\ln (\sec \theta )$.
We know that secant is an inverse function of cosine. So writing $\sec \theta $ as,
$\sec \theta = \dfrac{1}{{\cos \theta }}$
Hence we have,
$\ln (\sec \theta ) = \ln \left( {\dfrac{1}{{\cos \theta }}} \right)$
Where $\ln $ represents the natural logarithmic function to the base $e$.
Now we make use of properties of logarithmic function to simplify.
Here we use division property of logarithmic function given by,
$ \Rightarrow \ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b$
Here $a = 1$ and $b = \cos \theta $.
Substituting the values of a and b we get,
$\ln \left( {\dfrac{1}{{\cos \theta }}} \right) = \ln (1) - \ln (\cos \theta )$
We know that $\ln (1) = 0$.
Hence the above equation becomes,
$ \Rightarrow \ln \left( {\dfrac{1}{{\cos \theta }}} \right) = 0 - \ln (\cos \theta )$
$ \Rightarrow \ln \left( {\dfrac{1}{{\cos \theta }}} \right) = - \ln (\cos \theta )$
Which is the required right hand side.
Hence we have proved that the left hand side is equal to the right hand side.
Therefore, we have $\ln (\sec \theta ) = - \ln (\cos \theta )$.
Note: We must know the some basic trigonometric functions such as,
$\sec x = \dfrac{1}{{\cos x}}$, $\cos ecx = \dfrac{1}{{\sin x}}$, $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
If the question has the word log or $\ln $, it represents the given function as a logarithmic function. Note that we have two types of logarithmic function.
One is a common logarithmic function which is represented as a log and its base is 10.
The other one is a natural logarithmic function represented as $\ln $ and its base is $e$.
Some properties of logarithmic functions are given below.
(1) $\ln (x \cdot y) = \ln x + \ln y$
(2) $\ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y$
(3) $\ln {x^n} = n\ln x$
(4) $\ln 1 = 0$
(5) ${\log _e}e = 1$
So firstly consider the left hand side $\ln (\sec \theta )$. We know that secant is the inverse function of cosine function. So write $\sec \theta $ in terms of $\cos \theta $. Then we make use of properties of logarithm. In this we use division property given by $\ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b$. Substitute for the values of a and b to simplify and obtain the result which must be equal to the right hand side.
Complete step by step solution:
Given an equation $\ln (\sec \theta ) = - \ln (\cos \theta )$ …… (1)
We are asked to prove the left hand side is equal to the right hand side.
So let us begin with the L.H.S.
Consider the L.H.S. given by $\ln (\sec \theta )$.
We know that secant is an inverse function of cosine. So writing $\sec \theta $ as,
$\sec \theta = \dfrac{1}{{\cos \theta }}$
Hence we have,
$\ln (\sec \theta ) = \ln \left( {\dfrac{1}{{\cos \theta }}} \right)$
Where $\ln $ represents the natural logarithmic function to the base $e$.
Now we make use of properties of logarithmic function to simplify.
Here we use division property of logarithmic function given by,
$ \Rightarrow \ln \left( {\dfrac{a}{b}} \right) = \ln a - \ln b$
Here $a = 1$ and $b = \cos \theta $.
Substituting the values of a and b we get,
$\ln \left( {\dfrac{1}{{\cos \theta }}} \right) = \ln (1) - \ln (\cos \theta )$
We know that $\ln (1) = 0$.
Hence the above equation becomes,
$ \Rightarrow \ln \left( {\dfrac{1}{{\cos \theta }}} \right) = 0 - \ln (\cos \theta )$
$ \Rightarrow \ln \left( {\dfrac{1}{{\cos \theta }}} \right) = - \ln (\cos \theta )$
Which is the required right hand side.
Hence we have proved that the left hand side is equal to the right hand side.
Therefore, we have $\ln (\sec \theta ) = - \ln (\cos \theta )$.
Note: We must know the some basic trigonometric functions such as,
$\sec x = \dfrac{1}{{\cos x}}$, $\cos ecx = \dfrac{1}{{\sin x}}$, $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
If the question has the word log or $\ln $, it represents the given function as a logarithmic function. Note that we have two types of logarithmic function.
One is a common logarithmic function which is represented as a log and its base is 10.
The other one is a natural logarithmic function represented as $\ln $ and its base is $e$.
Some properties of logarithmic functions are given below.
(1) $\ln (x \cdot y) = \ln x + \ln y$
(2) $\ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y$
(3) $\ln {x^n} = n\ln x$
(4) $\ln 1 = 0$
(5) ${\log _e}e = 1$
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