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.