Write #secx+tanx# in terms of a tangent function?

2 Answers
Shwetank Mauria
Jul 11, 2017

#secx+tanx=tan(pi/4+x/2)#

Explanation:

#secx+tanx#

= #1/cosx+sinx/cosx#

= #(1+sinx)/cosx#

= #(sin^2(x/2)+cos^2(x/2)+2sin(x/2)cos(x/2))/(cos^2(x/2)-sin^2(x/2))#

= #(sin(x/2)+cos(x/2))^2/((cos(x/2)+sin(x/2))(cos(x/2)-sin(x/2)))#

= #(cos(x/2)+sin(x/2))/(cos(x/2)-sin(x/2))#

= #(1+sin(x/2)/cos(x/2))/(1-sin(x/2)/cos(x/2))#

= #(1+tan(x/2))/(1-tan(x/2))#

= #(tan(pi/4)+tan(x/2))/(1-tan(pi/4)tan(x/2))#

= #tan(pi/4+x/2)#

John D.
Jul 11, 2017

#secx+tanx = tan(x/2+pi/4)#

(I apologize in advance for the long explanation)

Explanation:

So we're looking for a way to represent #secx+tanx# in terms of a tangent function with the input manipulated. Let's start by looking at this rule:

#tan(a+b) = (tana+tanb)/(1-tanatanb)#

We'll need this later, so it's important to be able to recognize if an expression of this form comes up.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let's also look at the double angle formulas for cosine and tangent:

#cos(2x) = cos^2x-sin^2x#

#tan(2x) = (2tanx)/(1-tan^2x)#

We can replace #x# with #x/2# to get these handy formulas:

#cos(x) = cos^2(x/2) - sin^2(x/2)#

#tan(x) = (2tan(x/2))/(1-tan^2(x/2))#

Now, back to the original problem...

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let's write #secx# as #1/cosx# so we can use the formula we just made.

#secx+tanx = 1/cosx + tanx#

#= 1/(cos^2(x/2)-sin^2(x/2)) + (2tan(x/2))/(1-tan^2(x/2))#

Now we can divide both sides of the first fraction by #cos^2(x/2)#:

#= (1/cos^2(x/2))/((cos^2(x/2)-sin^2(x/2))/cos^2(x/2)) + (2tan(x/2))/(1-tan^2(x/2))#

#= sec^2(x/2)/(1-tan^2(x/2)) + (2tan(x/2))/(1-tan^2(x/2))#

Now both denominators are the same! Next, we should get rid of that #sec^2(x/2)# by using the rule #sec^2(a) = tan^2(a) + 1#

#= (tan^2(x/2)+1)/(1-tan^2(x/2)) + (2tan(x/2))/(1-tan^2(x/2))#

Now we can combine both fractions:

#= (tan^2(x/2)+1+2tan(x/2))/(1-tan^2(x/2)#

Hmm... these expressions look familiar. Remember these?

#a^2+2a+1 = (a+1)^2#

#1 - a^2 = (1-a)(1+a)#

Well, we can see that #a = tan(x/2)#, and that means we can factor both the top and the bottom of this fraction!

#= (tan(x/2)+1)^2/((1+tan(x/2))(1-tan(x/2)))#

#= (tan(x/2)+1)/(1-tan(x/2))#

We're almost done. This looks pretty close to the tangent angle addition formula. We only need one last thing: to turn #1# into a tangent expression. We can do that with this identity:

#tan(pi/4) = 1#

So we can substitute this into our expression like this:

#= (tan(x/2)+(1))/(1-(1)(tan(x/2)))#

#= (tan(x/2)+tan(pi/4))/(1-tan(pi/4)tan(x/2)#

This perfectly matches our angle addition formula! So now we can say that:

#(tan(x/2)+tan(pi/4))/(1-tan(pi/4)tan(x/2)) = tan(x/2+pi/4)#

And therefore #secx+tanx = tan(x/2+pi/4)#

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