Question

Write $\sec x + \tan x$ in terms of a tangent function.

Answer 1

Hint : We will first convert the given expression in terms of sine and cosine function. Then we will use the concept of half angle and the basic formula for ${\left( {a + b} \right)^2}$. Further we will simplify by using the formula of $\left( {{a^2} - {b^2}} \right)$. Then in order to convert the resulting expression into tangent function we will divide by cosine function and then try to convert the given expression into the formula for $\tan \left( {a + b} \right)$.

Complete step-by-step answer :
Given expression;
$\sec x + \tan x$
Now we will convert the given expression in terms of sine and cosine function.
We know,
$\sec x = \dfrac{1}{{\cos x}}$ and $\tan x = \dfrac{{\sin x}}{{\cos x}}$
So, the given expression becomes;
$= \dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}$
Now we will add the two fractions using $\cos x$ as the LCM. So, we get;
$= \dfrac{{1 + \sin x}}{{\cos x}}$
We know, $\sin 2x = 2\sin x\cos x$
So, $\sin 2\left( {\dfrac{x}{2}} \right) = 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$
Putting these values in the above expression we get,
$= \dfrac{{1 + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{\cos x}}$
Now we know that, $\cos x = {\cos ^2}\dfrac{x}{2} - {\sin ^2}\dfrac{x}{2}$. So, we will use the in the denominator. So, we get
$= \dfrac{{1 + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}$
Now we know that, ${\sin ^2}\dfrac{x}{2} + {\cos ^2}\dfrac{x}{2} = 1$, so we will replace $1$ in the numerator using this formula. So, we get;
$= \dfrac{{{{\sin }^2}x + {{\cos }^2}x + 2\cos \dfrac{x}{2}\sin \dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}$
We can see that the numerator can be written as ${\left( {a + b} \right)^2}$ and in the denominator we will use the formula; ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, so, we will get;
$= \dfrac{{{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}^2}}}{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}}$
Now we will cancel the common terms in the numerator and the denominator. We get,
$= \dfrac{{\left( {\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)}}{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)}}$
We have to get the result in terms of tangent function, so we will divide both the numerator and the denominator by $\cos \dfrac{x}{2}$. So, we get;
$= \dfrac{{\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + \dfrac{{\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}{{\dfrac{{\cos \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} - \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}$
Now we will use the formula that; $\tan x = \dfrac{{\sin x}}{{\cos x}}$. So, on further simplification we get,
$= \dfrac{{\tan \dfrac{x}{2} + 1}}{{1 - \tan \dfrac{x}{2}}}$
Now we will replace $1$, in the numerator by $\tan \dfrac{\pi }{4}$ because $\tan \dfrac{\pi }{4} = 1$.
$= \dfrac{{\tan \dfrac{x}{2} + \tan \dfrac{\pi }{4}}}{{1 - \tan \dfrac{x}{2}}}$
Now we can write the denominator as $\left( {1 - \tan \dfrac{x}{2}} \right) = \left( {1 - \tan \dfrac{\pi }{4}\tan \dfrac{x}{2}} \right)$. So, we get;
$= \dfrac{{\tan \dfrac{x}{2} + \tan \dfrac{\pi }{4}}}{{1 - \tan \dfrac{\pi }{4}\tan \dfrac{x}{2}}}$
As we can see that this is the formula for $\tan \left( {a + b} \right)$, $a = \dfrac{x}{2},b = \dfrac{\pi }{4}$. So, using this we get;
$= \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)$.
So, the correct answer is “$= \tan \left( {\dfrac{x}{2} + \dfrac{\pi }{4}} \right)$”.

Note : One major problem the students face here is that they get confused whether $\dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$ is $\tan \left( {a + b} \right)$ or $\tan \left( {a - b} \right)$. So, we should note here that while writing the formula for $\tan \left( {a + b} \right)$ we write plus in the numerator and minus in the denominator. While writing the formula for $\tan \left( {a - b} \right)$ we write minus in the numerator and plus in the denominator.