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-   -   what is the formula for Tan (A)-Tan (B) ? (http://mymathforum.com/calculus/19703-what-formula-tan-tan-b.html)

 silvercats June 6th, 2011 08:42 AM

what is the formula for Tan (A)-Tan (B) ?

what is the formula for Tan (A)-Tan (B) and Cot (A) - Cot(B)

 soroban June 6th, 2011 09:01 AM

Re: what is the formula for Tan (A)-Tan (B) ?

Hello, silvercats!

Quote:
 $\text{What is the formula for: }\:\tan A\,-\,\tan B\:\text{ and }\:\cot A \,-\,\cot B$

$\tan A \,-\,\tan B \;=\;\frac{\sin A}{\cos A}\,-\,\frac{\sin B}{\cos B} \;=\;\frac{\sin A\cos B\,-\,\cos A\sin B}{\cos A\,\cos B} \;=\;\frac{\sin(A\,-\,B)}{\cos A\,\cos B}$

$\cot A\,-\,\cot B \;=\;\frac{\cos A}{\sin A}\,-\,\frac{\cos B}{\sin B} \;=\;\frac{\sin B\cos A\,-\,\cos B\sin A}{\sin A\,\sin B} \;=\;\frac{\sin(B\,-\,A)}{\sin A\.\sin B}$

Are these what you're looking for?

 silvercats June 6th, 2011 05:56 PM

Re: what is the formula for Tan (A)-Tan (B) ?

Quote:
 Originally Posted by soroban Hello, silvercats! $\tan A \,-\,\tan B \;=\;\frac{\sin A}{\cos A}\,-\,\frac{\sin B}{\cos B} \;=\;\frac{\sin A\cos B\,-\,\cos A\sin B}{\cos A\,\cos B} \;=\;\frac{\sin(A\,-\,B)}{\cos A\,\cos B}$ $\cot A\,-\,\cot B \;=\;\frac{\cos A}{\sin A}\,-\,\frac{\cos B}{\sin B} \;=\;\frac{\sin B\cos A\,-\,\cos B\sin A}{\sin A\,\sin B} \;=\;\frac{\sin(B\,-\,A)}{\sin A\.\sin B}$ Are these what you're looking for?
I think ,NO. Like Cos(A)+Cos(B)=2 Cos[(A+B)/2].Cos [(A-B)/2],there is a formula for Tan too.not just sin/cos thing

 johnny June 6th, 2011 07:34 PM

tan A - tan B = (1 + tan A tan B)tan(A - B)

 silvercats June 7th, 2011 01:10 AM

Re: what is the formula for Tan (A)-Tan (B) ?

That is what I as looking for.Thanks for both of you :)

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