Question

How do you simplify $\dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}}$?

Answer 1

Hint : This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know the basic formula and conditions in trigonometric operations. Also, we need to know the arithmetic operations with the involvement of fraction terms. We need to know how to convert the mixed fraction terms into simple fraction terms.

Complete step-by-step answer :
The given expression is shown below,
$\dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = ? \to \left( 1 \right)$
We know that
$\tan x$Can also be written as,
$\tan x = \dfrac{{\sin x}}{{\cos x}} \to \left( 2 \right)$
And $\sec x$ can also be written as,
$\sec x = \dfrac{1}{{\cos x}}$ $\to \left( 3 \right)$
Let’s substitute the equation $\left( 2 \right)$and $\left( 3 \right)$ in the equation $\left( 1 \right)$, we get
$\left( 1 \right) \to \dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = ?$
$\dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = \dfrac{{\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{1}{{\cos x}}} \right)}} \to \left( 4 \right)$
Let’s solve the numerator part,
In the numerator we have,
$1 + \dfrac{{\sin x}}{{\cos x}} = ?$
We know that,
$a + \dfrac{b}{c} = \dfrac{{ac + b}}{c}$
By using this formula we can write,
$1 + \dfrac{{\sin x}}{{\cos x}} = \dfrac{{1 \cdot \cos x + \sin x}}{{\cos x}} = \dfrac{{\cos x + \sin x}}{{\cos x}}$
By using these values in the equation $\left( 4 \right)$, we get
$\left( 4 \right) \to \dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = \dfrac{{\left( {1 + \dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{1}{{\cos x}}} \right)}}$
$\dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = \dfrac{{\left( {\dfrac{{\cos x + \sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{1}{{\cos x}}} \right)}}$
The above equation can also be written as by using the formula
$\dfrac{{\left( {\dfrac{a}{b}} \right)}}{{\left( {\dfrac{d}{c}} \right)}} = \dfrac{a}{b} \cdot \dfrac{d}{c}$
We get
$\dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = \dfrac{{\left( {\dfrac{{\cos x + \sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{1}{{\cos x}}} \right)}} = \dfrac{{\cos x + \sin x}}{{\cos x}} \cdot \dfrac{{\cos x}}{1}$
By solving the above equation we get,
$\dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = \dfrac{{\cos x + \sin x}}{{\cos x}} \cdot \dfrac{{\cos x}}{1} = \dfrac{{\cos x + \sin x}}{1}$
So, the final answer is,
$\dfrac{{1 + \tan \left( x \right)}}{{\sec \left( x \right)}} = \cos x + \sin x$
So, the correct answer is “$\cos x + \sin x$”.

Note : Note that $\sin x$is the inverse form of $\csc x$. $\cos x$is the inverse form of $\sec x$. $\tan x$is the inverse form of $\cot x$. Also, note that $\tan x$can be written as $\dfrac{{\sin x}}{{\cos x}}$. Also, remember the basic formulae and conditions in trigonometric operations to make easy calculations. Also, note that if the expression is in the form of $\dfrac{{\left( {\dfrac{a}{b}} \right)}}{{\left( {\dfrac{d}{c}} \right)}}$it can be written as $\dfrac{a}{b} \cdot \dfrac{d}{c}$. Remember the basic algebraic formulae to solve these types of problems. Also, note that when we move one term from LHS to RHS or RHS to LHS, the arithmetic operations can be modified as follows,
Addition $\to$subtraction
Subtraction $\to$addition
Multiplication $\to$division
Division $\to$multiplication