# Trigo Identity

• Sep 9th 2011, 06:12 AM
janice91
Trigo Identity
Dear experts,

I am practicing the Trig identity, however met some issue in the changing $\displaystyle {\2tanB/(1+tanB)} TO {\1-3tan^2 B}$

identify: $\displaystyle \tan3B= \dfrac{\3tanB - tan^2 B }{\1-3tan^2 B}LHS := \dfrac{\ tanB + tan2 B }{\1-tanB.tan2B} = \dfrac{\3tanB - tan^2 B }{\2tanB/(1+tanB)}$
Thank you so much!
• Sep 9th 2011, 07:26 AM
Sudharaka
Re: Trigo Identity
Quote:

Originally Posted by janice91
Dear experts,

I am practicing the Trig identity, however met some issue in the changing $\displaystyle {\2tanB/(1+tanB)} TO {\1-3tan^2 B}$

identify: $\displaystyle \tan3B= \dfrac{\3tanB - tan^2 B }{\1-3tan^2 B}LHS := \dfrac{\ tanB + tan2 B }{\1-tanB.tan2B} = \boxed{\dfrac{\3tanB - tan^2 B }{\2tanB/(1+tanB)}}$

The boxed part is incorrect. Apply the double angle formula again for tan(2B) and you would get the answer.

Thank you so much!

.
• Sep 9th 2011, 10:31 AM
Soroban
Re: Trigo Identity
Hello, janice91!

Quote:

I met some issue in the changing .$\displaystyle \frac{2\tan B}{1+\tan B} \,\text{ to }\,1-3\tan^2B$

No wonder! . . . The two expressions are NOT equal.

• Sep 9th 2011, 08:44 PM
janice91
Re: Trigo Identity
I have made a careless mistake. Thank you for guidance!(Nod)