sum/difference identity for finding exact value of tan(-75°)

[Guest]

sum/difference identity for finding exact value of tan(-75°)

given tan(-75*), how would you figure out the exact value of it by using a sum or difference identity?

I only got as far as :
-tan(45+30) = (tan45+tan30) / (1-tan45+tan30)

 

[pka]

\(\displaystyle \L
\tan (45^o ) = 1\quad \& \quad \tan (30^o ) = \frac{{\sqrt 3 }}{3}\)

BTW: \(\displaystyle \L
\tan (A + B) = \frac{{\tan (A) + \tan (B)}}{{1 - \tan (A)\tan (B)}}\)

 

[Guest]

how do you simplify that further? i'm having trouble with the algebra part of it.

 

[soroban]

Re: sum/difference identity

Hello, amandamandy!

You were on the right track . . .

Given: \(\displaystyle \tan(-75^o)\)
Find the exact value by using a sum/difference identity.

The identity is: \(\displaystyle \L\,\tan(A\,+\,B)\;=\;\frac{\tan A\,+\,\tan B}{1\,-\,\tan A\cdot\tan B}\)


We have: \(\displaystyle \,\tan(-75^o)\;=\;-\tan(75^o)\;=\;-\tan(45\,+\,30)\)

\(\displaystyle \L\;\;\;=\;-\frac{\tan45\,+\,\tan30}{1\,-\,\tan45\cdot\tan30} \;=\;-\frac{1\,+\,\frac{1}{\sqrt{3}}}{1\,-\,\frac{1}{\sqrt{3}}} \;=\;-\frac{\sqrt{3}\,+\,1}{\sqrt{3}\,-\,1}\)

Rationalize: \(\displaystyle \L\,-\frac{\sqrt{3}\,+\,1}{\sqrt{3}\,-\,1}\cdot\frac{\sqrt{3}\,+\,1}{\sqrt{3}\,+\,1}\;=\;\frac{3\,+\,2\sqrt{3}\,+\,1}{3\,-\,1}\;=\;-\frac{4\,+\,2\sqrt{3}}{2}\;=\;-2\,-\,\sqrt{3}\)