Trigo Identity

**janice91** · September 9th 2011, 06:12 AM

Dear experts,

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

identify: $\tan3B= \dfrac{\3tanB - tan^2 B }{\1-3tan^2 B}LHS := \dfrac{\ tanB + tan2 B }{\1-tanB.tan2B} = \dfrac{\3tanB - tan^2 B }{\2tanB/(1+tanB)}$
Thank you so much!

**Sudharaka** · September 9th 2011, 07:26 AM

Originally Posted by janice91

Dear experts,

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

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

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

Thank you so much!

.

**Soroban** · September 9th 2011, 10:31 AM

Hello, janice91!

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

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

**janice91** · September 9th 2011, 08:44 PM

I have made a careless mistake. Thank you for guidance!

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