Question #b9e87

2 Answers
Tony B
May 4, 2017

It all depends on where you wish to take this. One option is:

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

Explanation:

Tony B

SohCahToa

Soh#-> sin(x)=("opposite")/("hypotenuse") ->b/a#

Cah#->cos(x)=("adjacent")/("hypotenuse")->c/a#

Toa#->tan(x)=("opposite")/("adjacent")->b/c#

#tan(x)=(sin(x))/(cos(x))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let the unknown term be #beta#

Given equation: #(tan(x))/(cos(x))=beta" ".........Equation(1)#

Write as: #" "tan(x)xx1/(cos(x))=beta#

But #tan(x)=(sin(x))/(cos(x))# so by substitution we have:

#" "(sin(x))/(cos(x))xx1/(cos(x))=beta#

#" "sin(x)/(cos(x))^2=beta#

This is written as:

#" "sin(x)/cos^2(x)=beta" "...............Equation(2)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As in the Pythagoras equation #a^2=b^2+c^2#
There is a comparable one involving #cos^2(x)#

#"Pythagorean relationship"->sin^2(x)+cos^2(x)=1#

So #" "cos^2(x)=1-sin^2(x)" ".............Equation(3)#

Using #Equation(3)# substitute for #cos^2(x)# in #Equation(2)#

#" "sin(x)/cos^2(x)=beta" "->" "sin(x)/(1-sin^2(x))=beta#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Thus the finished relationship is:

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

Walker P
May 4, 2017

#sin(x)/cos^2(x)# or #sin(x)/(1-sin^2(x))#

Explanation:

First of all, what are #tan(x)# and #cos(x)#?

Both are trigonometric functions, but #tan(x)# is actually a function of both #sin(x)# and #cos(x)#:

#tan(x)=sin(x)/cos(x)#

Substitute this in for the numerator in the original problem:

#tan(x)/cos(x)=(sin(x)/cos(x))/cos(x)#

Remember that dividing by anything is the same as multiplying by the reciprocal of that thing. In other words, the following is mathematically valid:

#a/b=a*1/b#

Applying this to our equation, we see that:

#(sin(x)/cos(x))/cos(x)=sin(x)/cos(x)*1/cos(x)#

Simplifying, we get:
#sin(x)/cos(x)*1/cos(x)=sin(x)/cos^2(x)#

If we want our answer to be in terms of #sin(x)#, we could go further and use the identity: #cos^2(x)+sin^2(x)=1#

Replacing #cos^2(x)#, we get:

#sin(x)/(1-sin^2(x))#

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