Proof

H

HopeWelch

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Oct 31, 2011
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I need to prove that (1+tan^3X)/(1+tanX)=1-tanX+tan^2X.
I am only allowed to manipulate the left side of the equation.
I had 25 problems to do for homework and this is the only one I don't understand. Did all other 24 just fine.

So far I have tried substituting in all sines and cosine but that doesn't seem to get me anywhere and I've tried multiplying by 1 in the forms (1-tanX)/(1-tanX) and (1+tanX)/(1+tanX). They just turns into a bunch of terms separated by + and - that do not seem to be going anywhere or canceling. I'm probably just missing something simple. Please help.
 
HopeWelch said:
I need to prove that (1+tan^3X)/(1+tanX)=1-tanX+tan^2X.
I am only allowed to manipulate the left side of the equation.
I had 25 problems to do for homework and this is the only one I don't understand. Did all other 24 just fine.

So far I have tried substituting in all sines and cosine but that doesn't seem to get me anywhere and I've tried multiplying by 1 in the forms (1-tanX)/(1-tanX) and (1+tanX)/(1+tanX). They just turns into a bunch of terms separated by + and - that do not seem to be going anywhere or canceling. I'm probably just missing something simple. Please help.
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Here's a place where you can make good use of one of those factoring patterns you learned in algebra.

Factoring a SUM of two cubes: \(\displaystyle a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

And....

Factoring a DIFFERENCE of two cubes: \(\displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)

The NUMERATOR on the left side is

\(\displaystyle 1 + tan^3 x\)

If you think about it for a minute, that's a SUM of two cubes....

\(\displaystyle 1^3 + (tan x)^3\)

Apply the appropriate factoring pattern, and reduce the fraction. I think you'll be pleased.
 
Thanks for your reply Mrspi. I took a long break between functions and trigonometry and totally forgot about the sum of cubes. I remember now. Knew I was missing something simple. Thank you again!!:D
 
And remember the acronym S.O.A.P for remembering the signs in the factoring of the sum and difference of two cubes.

S = Same \(\displaystyle a^3+b^3=(a\Rightarrow+\Leftarrow b)(a^2-ab+b^2)\)
O = Opposite \(\displaystyle a^3+b^3=(a+b)(a^2\Rightarrow-\Leftarrow ab+b^2)\)
A.P. = Always Positive \(\displaystyle a^3+b^3=(a+b)(a^2-ab\Rightarrow+\Leftarrow b^2)\)

Similarly: \(\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)\)
 
S.o.a.p.

Nice acronym Srmichael. I will never forget this again!

Also, I'm new to this forum today and so impressed. Thank you all for the speedy replies and the tips!
 
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