Question

How do I simplify sec x by tan x?

Answer 1

Hint: To simplify $ \dfrac{{\sec x}}{{\tan x}} $ , we need to look for the trigonometric relations and functions that can be applicable here. Now, we know that sec is inverse of cosine and tan is the ratio of sine and cosine. So, 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{{\sec x}}{{\tan x}} $ .
Now, we need to simplify this.
Here, we have sec x in numerator and tan x in denominator. To simplify this, we simply need to look for some relations or formulas that are applicable here.
Now, we know that sec is the inverse of cosine, so one of the relations that can be used is that we can write sec as 1 divided by cos.
 $ \to \sec x = \dfrac{1}{{\cos x}} $
Now, for tan we know that tan is inverse of cot, so we can write tan as 1 divided by cot.
 $ \to \tan x = \dfrac{1}{{\cot x}} $
But there is no relation between cos and cot, so we need to find some other relation.
Another relation for tan is than it is the ratio of sin and cos. So, we can write tan as sin divided by cos.
 $ \to \tan x = \dfrac{{\sin x}}{{\cos x}} $
We could use this relation in our question.
Hence, using this relations, our expression will become
 $ \to \dfrac{{\sec x}}{{\tan x}} = \dfrac{{\dfrac{1}{{\cos x}}}}{{\dfrac{{\sin x}}{{\cos x}}}} $
Now, the cos term in numerator will go in denominator and the cos term in the denominator will go in numerator.
 $ \to \dfrac{{\sec x}}{{\tan x}} = \dfrac{{1 \times \cos x}}{{\sin x \times \cos x}} $
Cancelling cos x, we get
\[ \to \dfrac{{\sec x}}{{\tan x}} = \dfrac{1}{{\sin x}}\]
Now, the inverse of sine is cosine. Therefore,
\[ \to \dfrac{{\sec x}}{{\tan x}} = \cos ecx\]
Hence, on simplifying $ \dfrac{{\sec x}}{{\tan x}} $ we get \[\cos ecx\].
So, the correct answer is “Option B”.

Note: We can also simplify this using other method.
Given: $ \dfrac{{\sec x}}{{\tan x}} $ .
Now, we know that $ \sec x = \dfrac{{hypotenuse}}{{adjacent}} $ and $ \tan x = \dfrac{{opposite}}{{adjacent}} $ .
Putting this values in our expression, we get
 $ \to \dfrac{{\sec x}}{{\tan x}} = \dfrac{{\dfrac{{hypotenuse}}{{adjacent}}}}{{\dfrac{{opposite}}{{adjacent}}}} $
Here, adjacent gets cancelled. Therefore,
 $ \to \dfrac{{\sec x}}{{\tan x}} = \dfrac{{hypotenuse}}{{opposite}} $
Now, we know that $ \dfrac{{hypotenuse}}{{opposite}} = \cos ecx $ . Therefore,
 $ \to \dfrac{{\sec x}}{{\tan x}} = \cos ecx $