Develop the identity for tan(A - B) by using the sine and co

[NEHA]

Develop the identity for tan(A - B) by using the sine and cosine difference identities.

so i have to use this way :

Tan(x-y) = tanx - tany / 1 + tan x tan y

http://www.mathwords.com/t/trig_identities.htm

so
tan(A - B) = tanA - tanB / 1 + tanA * tanB
thats it ..because check that site thats all that the site tells . so this has ot be it .

 

[stapel]

Re: Develop the identity for tan(A - B) by using the sine an

thats it ..because check that site thats all that the site tells . so this has ot be it .

I'm pretty sure the exercise wants you to start by converting the tangent to sines and cosines, and then work with sine and cosine identities to "develop" the tangent identity. Your book almost certainly contains the identity. The point isn't to find the identity (in your book or from some other source), but to "develop" it yourself.

I could be wrong, of course....

Eliz.

 

[NEHA]

I GOT IT

Tan(A-B) =

SinA CosB/CosA CosB - CosA SinB/CosA CosB
------------------------------------------------------
CosA CosB/CosA CosB + SinA SinB/CosA CosB


and

Tan (A-B) =

TanA - TanB
------------------
1 + TanA TanB

 

[soroban]

Re: Develop the identity for tan(A - B) by using the sine an

Hello, NEHA!

Develop the identity for \(\displaystyle \tan(A\,-\,B)\)
by using the sine and cosine difference identities.

They want you to develop the identity . . . not copy it from a book!


Start with: \(\displaystyle \L\;\tan(A\,-\,B)\;=\;\frac{\sin(A\,-\,B)}{\cos(A\,-\,B)}\;\) [1]


The sine and cosine difference formulas are:
. . \(\displaystyle \sin(A\,-\,B)\;=\;\sin(A)\cdot\cos(B)\,-\,\sin(B)\cdot\cos(A)\)
. . \(\displaystyle \cos(A\,-\,B)\;=\;\cos(A)\cdot\cos(B)\,+\,\sin(A)\cdot\cos(B)\)


Substitute these into [1]:

\(\displaystyle \L\tan(A\,-\,B) \;=\;\frac{\sin(A)\cdot\cos(B)\,-\,\sin(B)\cdot\cos(A)}{\cos(A)\cdot\cos(B)\,+\,\sin(A)\cdot\sin(B)}\)



Divide top and bottom by \(\displaystyle \cos(A)\cdot\cos(B):\)

\(\displaystyle \L\tan(A\,-\,B)\;=\;\frac{\frac{\sin(A)\cdot\sout{\cos(B)}}{\cos(A)\cdot\sout{\cos(B)}}\,-\,\frac{\sin(B)\cdot\sout{\cos(A)}}{\sout{\cos(A)}\cdot\cos(B)}} {\frac{\sout{\cos(A)}\cdot\sout{\cos(B)}}{\sout{\cos(A)}\cdot\sout{\cos(B)}} \,+\,\frac{\sin(A)\cdot\sin(B)}{\cos(A)\cdot\cos(B)} }\)

\(\displaystyle \L\tan(A\,-\,B)\;=\;\frac{\frac{\sin(A)}{\cos(A)}\,-\,\frac{\sin(B)}{\cos(B)} }{1\,+\,\frac{\sin(A)}{\cos(A)}\cdot\frac{\sin(B)}{\cos(B)}}\)

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