As written, your denominator is zero. Fix the error here:
View attachment 30614
i think i added parenthesis already? where else sir?Check your signs in the denominator starting with the first equal sign.
The second line to last, please add parenthesis. Without them, it means something different.
Cos(A + B) = Cos(A) * Cos(B) - Sin(A) * Sin(B)i do not grab that, is it supposed to be - ?
The work, as well as the answer, looks correct to me.What about this method?
... which was corrected on the next line, so I didn't bother looking up. Such is the life of a proofreader.
much thanks
so it should be enclosed as (2cosA.cosB)/(2sinA.sinB) ?Yes. It's supposed to be "-". You have - for the same identity in the numerator.
View attachment 30615
This line is missing a parenthesis. What you wrote means [math]\frac{(2\cos A \cos B)(\sin A\sin B)}{2} \neq\frac{2\cos A \cos B}{2\sin A\sin B}[/math]
Correctmuch thanks
so it should be enclosed as (2cosA.cosB)/(2sinA.sinB) ?
But after collecting like terms, the sign has to changeThe signs are still wrong but somehow got the right answer at the end.
It should not change after collecting like terms?
Signs do not change because of collecting like terms. You didn't distribute the negative sign properly in the picture I posted in #11It should not change after collecting like terms?
Is the sign at that point just the only error which does not affect other setting?
Just to change + to - and that is all?