Question

How do you simplify $\cos x + \sin x\tan x?$

Answer 1

First convert all the terms present in the given expression in the form of sine and cosine $({\text{i}}{\text{.e}}{\text{.}}\;\sin \;{\text{and}}\;\cos )$ and then simplify further and try to implement or apply trigonometric identities related to sine and cosine functions and finally simplify it.

Following trigonometric identities will be helpful in this question:
$\tan x = \dfrac{{\sin x}}{{\cos x}}\;{\text{and}}\;\cos x = \dfrac{1}{{\sec x}}$
${\sin ^2}x + {\cos ^2}x = 1$

Complete step by step solution:
In order to simplify $\cos x + \sin x\tan x$ we will first change all the terms in the given trigonometric expression into sine and cosine.

In the expression $\cos x + \sin x\tan x$ there is a tangent $({\text{i}}{\text{.e}}{\text{.}}\;\tan x)$, we have to convert $\tan x$ into sine and cosine

We know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$, so putting $\dfrac{{\sin x}}{{\cos x}}$ in place of $\tan x$ in the given expression $\cos x + \sin x\tan x$, we will get
$= \cos x + \sin x\tan x \\\ = \cos x + \sin x \times \dfrac{{\sin x}}{{\cos x}} \\\ = \cos x + \dfrac{{{{\sin }^2}x}}{{\cos x}} \\\$
Taking LCM for simplifying further, we will get
$= \cos x + \dfrac{{{{\sin }^2}x}}{{\cos x}} \\\ = \dfrac{{\cos x \times \cos x + {{\sin }^2}x}}{{\cos x}} \\\ = \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\cos x}} \\\$
See the expression in the numerator of the simplified expression, are we familiar with this?

Yes, this is an trigonometric identity and it is given as
${\sin ^2}x + {\cos ^2}x = 1$

Using this identity in the expression to simplify further,

Writing $1$ in the place of ${\cos ^2}x + {\sin ^2}x$, we will get
$= \dfrac{{{{\cos }^2}x + {{\sin }^2}x}}{{\cos x}} \\\ = \dfrac{1}{{\cos x}} \\\$
Now we know that $\cos x = \dfrac{1}{{\sec x}} \Rightarrow \sec x = \dfrac{1}{{\cos x}}$ , therefore above expression can be further written as
$= \dfrac{1}{{\cos x}} \\\ = \sec x \\\$
Therefore the given expression $\cos x + \sin x\tan x$ is simplified to $\sec x$

Note: Cosecant, secant and cotangent (represented as $\sec ,\;\csc \;{\text{and}}\;\cot$) are multiplicative inverses of sine, cosine and tangent (represented as $\sin ,\;\cos \;{\text{and}}\;\tan$) respectively.

But at the same time, the functions cosecant, secant and cotangent are also not defined at some angles where sine, cosine and tangent are defined.

This is because where the sine is equals to $0$ its cosecant will be given as $\dfrac{1}{0}$ which is not defined, similar with the cosine-secant and
tangent-cotangent.