when tanA+tanB+tanC = tanAtanBtanC

If A+B+C = 180.

tanA=tan(180-B-C)

= \frac{tan(180-B)-tanC}{1+tan(180-B)tanC}

= \displaystyle \frac{\frac{tan180-tanB}{1+tan180tanB}-tanC}{1+\frac{tan180-tanB}{1+tan180tanB}tanC}

= \frac{-tanB-tanC}{1-tanBtanC}

Then, we have tanA(1-tanBtanC)=-tanB-tanC \blacksquare

Advertisements
This entry was posted in Uncategorized. Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s