Question

How do you rewrite the expression as a single logarithm and simplify $\ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right)$?

Answer 1

Hint: We will first of all use the fact that $\ln a + \ln b = \ln ab$. Then we will write both the cotangent and tangent in terms of sine and cosine and simplify the expression further to get the required simplification.

Complete step by step solution:
We are given that we are required to rewrite the expression as a single logarithm and simplify $\ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right)$.
Now, since we know that we have a formula given by the following expression with us regarding the logarithmic function:-
$ \Rightarrow \ln a + \ln b = \ln ab$
Now, let us replace a by |cot t| and b by $1 + {\tan ^2}t$, we will then obtain the following expression with us:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = \ln \left\{ {|\cot t|\left( {1 + {{\tan }^2}t} \right)} \right\}\]
Since, we know that 1 > 0 and \[{\tan ^2}t > 0\].
Adding both of these, we will then obtain the following inequality:-
\[ \Rightarrow 1 + {\tan ^2}t > 0\]
Therefore, we can write the earlier expression as follows:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = \ln \left\{ {|\cot t\left( {1 + {{\tan }^2}t} \right)|} \right\}\]
Re – writing the above expression as follows:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = \ln \left\{ {\left| {\dfrac{{\cos t}}{{\sin t}}\left( {1 + \dfrac{{{{\sin }^2}t}}{{{{\cos }^2}t}}} \right)} \right|} \right\}\]
Simplifying the right hand side of the above expression, we will then obtain the following expression with us:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = \ln \left\{ {\left| {\dfrac{{\cos t}}{{\sin t}} + \dfrac{{\cos t}}{{\sin t}} \times \dfrac{{{{\sin }^2}t}}{{{{\cos }^2}t}}} \right|} \right\}\]
Simplifying the right hand side of the above expression further, we will then obtain the following expression with us:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = \ln \left\{ {\left| {\dfrac{{\cos t}}{{\sin t}} + \dfrac{{\sin t}}{{\cos t}}} \right|} \right\}\]
Taking the least common multiple and then solving the right hand side, we will then obtain the following equation with us:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = \ln \left\{ {\left| {\dfrac{{{{\cos }^2}t + {{\sin }^2}t}}{{\sin t\cos t}}} \right|} \right\}\]
Since we know that we have an identity given by \[{\cos ^2}t + {\sin ^2}t = 1\]. Therefore, we will then obtain the following equation with us:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = \ln \left\{ {\left| {\dfrac{1}{{\sin t\cos t}}} \right|} \right\}\]
Since, we also know that we have a formula given by $\ln {a^{ - 1}} = - \ln a$. Therefore, we get:-
\[ \Rightarrow \ln |\cot t| + \ln \left( {1 + {{\tan }^2}t} \right) = - \ln \left| {\sin t\cos t} \right|\]

Note: The students must know that the logarithm of negative or zero is not defined, therefore, in the question, we have the modulus of cot t because we know that 1 > 0 and \[{\tan ^2}t > 0\].
Adding both of these, we will then obtain the following inequality:-
\[ \Rightarrow 1 + {\tan ^2}t > 0\]
The terms other than the cot t inside the logarithmic is already positive, therefore, we took the modulus of cot t.
\[ \Rightarrow |\cot t|\left( {1 + {{\tan }^2}t} \right) > 0\]
The students must commit to memory the following formulas we used in the solution above:-
1. $\ln {a^{ - 1}} = - \ln a$
2. \[{\cos ^2}t + {\sin ^2}t = 1\]
3. $\ln a + \ln b = \ln ab$
4. $\tan t = \dfrac{{\sin t}}{{\cos t}}$
5. $\cot t = \dfrac{{\cos t}}{{\sin t}}$