Question

How do you simplify the expression \[\dfrac{{\tan \theta }}{{\sec \theta }}\]?

Answer 1

Hint: To simplify \[\dfrac{{\tan \theta }}{{\sec \theta }}\], we need to find the trigonometric relations and formulas applicable here. Now, we know that \[\tan \] is the ratio of \[\sin e\] and \[\cos ine\] and \[\sec \] is the inverse of \[\cos ine\]. So, we will substitute these values in the given expression and we will get our expression simplified.

Complete step by step answer:
In this question, we are given a trigonometric ratio and we need to simplify it.
Given, \[\dfrac{{\tan \theta }}{{\sec \theta }}\].
Now, we need to simplify this.
Here, we have \[\tan \theta \] in the numerator and \[\sec \theta \] in the denominator. To simplify this, we simplify using some relations or formulas that are applicable here.
We know that \[\tan \] is the ratio of \[\sin e\] and \[\cos ine\]. So, we can write \[\tan \] as \[\sin \] divided by \[\cos \].
\[ \Rightarrow \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Now, we know that \[\sec \] is the inverse of \[\cos ine\], so one of the relations that can be used is that we can write \[\sec \] as \[1\] divided by \[\cos \].
\[ \Rightarrow \sec \theta = \dfrac{1}{{\cos \theta }}\]
We could use these two relations in our question.
Hence, using this relation, our expression will become
\[ \Rightarrow \dfrac{{\tan \theta }}{{\sec \theta }} = \dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{1}{{\cos \theta }}}}\]
Now, the \[\cos \] term in the numerator will go in the denominator and the \[\cos \] term in the denominator will go in the numerator. So, we get
\[ \Rightarrow \dfrac{{\tan \theta }}{{\sec \theta }} = \dfrac{{\sin \theta \times \cos \theta }}{{1 \times \cos \theta }}\]
Cancelling \[\cos \theta \], we get
\[ \Rightarrow \dfrac{{\tan \theta }}{{\sec \theta }} = \sin \theta \]
Therefore, by simplifying \[\dfrac{{\tan \theta }}{{\sec \theta }}\], we get \[\sin \theta \].

Note:
We can also simplify by another method.
Given, \[\dfrac{{\tan \theta }}{{\sec \theta }}\].
As we know, \[\tan \theta = \dfrac{{perpendicular}}{{base}}\] and \[\sec \theta = \dfrac{{hypotenuse}}{{base}}\].
Putting these values in our expression i.e., \[\dfrac{{\tan \theta }}{{\sec \theta }}\], we get
\[ \Rightarrow \dfrac{{\tan \theta }}{{\sec \theta }} = \dfrac{{\dfrac{{perpendicular}}{{base}}}}{{\dfrac{{hypotenuse}}{{base}}}}\]
On cancelling the common terms from the numerator and the denominator, we get
\[ \Rightarrow \dfrac{{\tan \theta }}{{\sec \theta }} = \dfrac{{perpendicular}}{{hypotenuse}}\]
We know that \[\dfrac{{perpendicular}}{{hypotenuse}} = \sin \theta \]. Therefore,
\[ \Rightarrow \dfrac{{\tan \theta }}{{\sec \theta }} = \sin \theta \]
Therefore, by simplifying \[\dfrac{{\tan \theta }}{{\sec \theta }}\], we get \[\sin \theta \].