trigonometric equation

**mi986** · October 3rd 2008, 10:16 AM

Hi, I have problem that asks, "Prove that the equations are identities."
I tried for hours to solve it, but I'm lost... I would really appreciate it if someone could help me on this.

cot A + tan A + 1 = (cot A)/(1-tan A) + (tan A)/(1-cot A)

The stipulation is that I can only use the trigonometric identities that I've learned so far..

Such as:
csc a = 1/sin a
sec a = 1/csc a
tan a = sin a/cos a
etc...

The pythgorean identities and Opposite angle formulas

So far I've tried switching it sines and cosines, tangent, and a few others but it never seems to come out right...

Thanks to anyone who can help.

**Jhevon** · October 3rd 2008, 10:22 AM

Originally Posted by mi986

Hi, I have problem that asks, "Prove that the equations are identities."
I tried for hours to solve it, but I'm lost... I would really appreciate it if someone could help me on this.

cot A + tan A + 1 = (cot A)/(1-tan A) + (tan A)/(1-cot A)

The stipulation is that I can only use the trigonometric identities that I've learned so far..

Such as:
csc a = 1/sin a
sec a = 1/csc a
tan a = sin a/cos a
etc...

The pythgorean identities and Opposite angle formulas

So far I've tried switching it sines and cosines, tangent, and a few others but it never seems to come out right...

Thanks to anyone who can help.

start by adding the fractions on the right hand side and simplify. now do you see how to get it?

**mi986** · October 3rd 2008, 10:33 AM

That's one of the first things I did. :-s

**Soroban** · October 3rd 2008, 11:32 AM

Hello, mi986!

Here's one way . .

$\cot A + \tan A + 1 \:= \:\frac{\cot A}{1-\tan A} + \frac{\tan A}{1-\cot A}$

The right side is: . $\frac{\frac{1}{\tan A}}{1 - \tan A} + \frac{\tan A}{1 - \frac{1}{\tan }}$

Multiply each fraction by $\frac{\tan A}{\tan A}\!:\quad {\color{blue}\frac{\tan A}{\tan A}}\cdot\frac{\frac{1}{\tan A}}{1 - \tan A} + {\color{blue}\frac{\tan A}{\tan A}} \cdot\frac{\tan A}{1 - \frac{1}{\tan A}}$

. . $=\; \frac{1}{\tan A(1 - \tan A)} + \frac{\tan^2A}{\tan A - 1} \;=\;\frac{1}{\tan A(1 - \tan A)} - \frac{\tan^2\!A}{1-\tan A}$

Multiply the second fraction by $\frac{\tan A}{\tan A}\!:\quad\frac{1}{\tan A(1 - \tan A)} - \frac{\tan^3\!A}{\tan A(1 - \tan A)}$

. . and we have: . $\frac{1-\tan^3\!A}{\tan (1 - \tan A)}\;\;^{\leftarrow\;\text{difference of cubes}}$

Factor: . $\frac{(1-\tan A)(1 + \tan A + \tan^2\!A)}{\tan A(1 - \tan A)} \;=\;\frac{1 + \tan A + \tan^2\!A}{\tan A}$

. . $= \;\frac{1}{\tan A} + \frac{\tan A}{\tan A} + \frac{\tan^2\!A}{\tan A} \;=\; \cot A + 1 + \tan A$

**mi986** · October 3rd 2008, 06:34 PM

Ooh! Thank you so much!

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