Prove: tan(A+45) = (1+tanA)/(1-tanA)
Please i need help, i know that you start with the RHS ... but i do not know.
Also try to work with only "sin" and "cos" as we havent learnt any other methods yet.
Thanks so much
Prove: tan(A+45) = (1+tanA)/(1-tanA)
Please i need help, i know that you start with the RHS ... but i do not know.
Also try to work with only "sin" and "cos" as we havent learnt any other methods yet.
Thanks so much
The only other way I can think to help you with this is to use the sum of angles formula for sine and cosine:
$\displaystyle sin(A + B) = sin(A)~cos(B) + sin(B)~cos(A)$
and
$\displaystyle cos(A + B) = cos(A)~cos(B) - sin(A)~sin(B)$
Then
$\displaystyle tan(A + 45) = \frac{sin(A + 45)}{cos(A + 45)}$
$\displaystyle = \frac{sin(A)~cos(45) + sin(45)~cos(A)}{cos(A)~cos(45) - sin(A)~sin(45)}$
which can be reduced to the needed expression.
I don't think that there is any way to do this problem without using one of these methods.
-Dan
The only thing I can think of to do is continue:
$\displaystyle = \frac{\frac{\sqrt{2}}{2}~sin(A) + \frac{\sqrt{2}}{2}~cos(A)}{\frac{\sqrt{2}}{2}~cos( A) - \frac{\sqrt{2}}{2}~sin(A)}$
$\displaystyle = \frac{sin(A) + cos(A)}{cos(A) - sin(A)}$
Now divide numerator and denominator by cos(A).
-Dan
I have spent the last 4 hours doing 15 or so trig "proving identities" and i really cannot get this one, even after u have said that, i am at the last stage with sina+cosA/cosa-sina...
please can you just show me in numbers, i aint good with reading numbers, as in divide numerator by x etc, i need to see it to understand ...thanks
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You don't need to use sin or cos anyway, the tan(A+B) formula constitutes a sufficient proof, and looking at the question, that is probably the one that you are supposed to use.
Start on the left and work to the right. Either use the sin and cosine addition formulae
sin(A+B)=sinAcosB+sinBcosA
cos(A+B)=cosAcosB-sinAsinB
and the identity tanx=sinx/cosx
which is the same as saying tan(A+B)=sin(A+B)/cos(A+B)
then the rest should be quite easy, just replace the sin and cos in the fraction with the identities above.
Alternatively and much easier, there is an additional identity,
tan(A+B)=tanA-tanB/1-tanAtanB
which you can take to be true (it's proved easily by expressing tan as sin/cos but you don't need to do this).