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Question: How do you rewrite the expression as a single log and simplify \[\ln \left| \sin \theta \right|-\ln ...

How do you rewrite the expression as a single log and simplify ln⁡∣sin⁡θ∣−ln⁡∣cos⁡θ∣\ln \left| \sin \theta \right|-\ln \left| \cos \theta \right| ?

Explanation

Solution

In the given question, we have been asked to write an equation as a function of a single log. In order to solve the question, we need to apply the trigonometric formula or identities to combine the angles. The formula we will use in the given question is sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin \theta }{\cos \theta }=\tan \theta . Here, we apply this formula after combining both the log terms by using the quotient rule of logarithm which states that the logarithm of the quotient is equal to the difference of logarithms. Then we will further simplify the given expression.

Formula used:
The quotient rule of logarithm which states that the logarithm of the quotient is equal to the difference of logarithms, i.e.
log⁡(AB)=log⁡A−log⁡B\log \left( \dfrac{A}{B} \right)=\log A-\log B
sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin \theta }{\cos \theta }=\tan \theta

Complete step by step answer:
We have given that,
⇒ln⁡∣sin⁡θ∣−ln⁡∣cos⁡θ∣\Rightarrow \ln \left| \sin \theta \right|-\ln \left| \cos \theta \right|
Using the quotient rule of logarithm which states that the logarithm of the quotient is equal to the difference of logarithms, i.e.
log⁡(AB)=log⁡A−log⁡B\log \left( \dfrac{A}{B} \right)=\log A-\log B
Applying the quotient rule of logarithm in the given expression, we get
⇒ln⁡∣sin⁡θ∣−ln⁡∣cos⁡θ∣=ln⁡(∣sin⁡θ∣∣cos⁡θ∣)\Rightarrow \ln \left| \sin \theta \right|-\ln \left| \cos \theta \right|=\ln \left( \dfrac{\left| \sin \theta \right|}{\left| \cos \theta \right|} \right)
Now, using the trigonometric identity, i.e. sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin \theta }{\cos \theta }=\tan \theta
Substituting the value of sin⁡θcos⁡θ=tan⁡θ\dfrac{\sin \theta }{\cos \theta }=\tan \theta in the above expression.Here, by taking the absolute value of both the sine and cos function, tan function is forced to be positive.Thus,
ln⁡(∣sin⁡θ∣∣cos⁡θ∣)=ln⁡(∣tan⁡θ∣)\ln \left( \dfrac{\left| \sin \theta \right|}{\left| \cos \theta \right|} \right)=\ln \left( \left| \tan \theta \right| \right)
Therefore, we get
∴ln⁡∣sin⁡θ∣−ln⁡∣cos⁡θ∣=ln⁡∣tan⁡θ∣\therefore\ln \left| \sin \theta \right|-\ln \left| \cos \theta \right|=\ln \left| \tan \theta \right|

Hence, it is the required answer.

Note: While solving these types of questions, students need to know the basic concepts of logarithm and the trigonometric functions. In order to solve the above given question, students need to remember the basic properties of logarithmic function to combine two log functions given in any expression. Applying the identities would make the question easier to solve.