Prove:
sinx + tanx = tanx (1 + cosx)
What I have so far:
LS:
= sinx + tanx
= sinx + (sinx / cosx)
= (sinx) (cosx) + sinx / cos
= tanx (cosx + sinx)
I don't know what to do now
Your second last line needs brackets, and the following line makes no sense
Why not work on the right side?
RS = tanx(1+cosx)
=tanx + tanxcosx after expanding
= tanx + (sinx/cosx)cosx
= tanx + sinx
= LS
My second last line:
(sinx) (cosx) + sinx
_________________________
cosx
When I did the right side, I get the same problem as the left side:
RS:
= tanx + (1 + cosx)
= (sinx / cosx) (1 + cosx)
= (1 + cosx) (cosx)
= (sinx) (1 + cosx)
_______________________
cosx
= (sinx) + (cosx)(sinx)
________________________
cosx
And ... my teacher wants everything in terms of cos and sin
RS = tanx(1+cosx)
=tanx + tanxcosx after expanding
= tanx + (sinx/cosx)cosx
= tanx + sinx
= LS
For this line "=tanx + tanxcosx after expanding", how did you get the second tanx from and what has happened to the addition sign?
from you second last line
(sinx) (cosx) + sinx
_________________________
cosx
= sinx[cosx + 1]/cosx , ....I took out a common factor
= tanx(cosx + 1)
= RS
You said "And ... my teacher wants everything in terms of cos and sin"
your teacher probably suggested changing everything to sines and cosines as you work your way through. Notice that is also what I did as I showed the steps.
I would never have restricted my students to use only a part of their trig knowledge, why not make use of all the relationships at your hand, picking the simplest way?
secx/sinx - sinx/cosx = cotx
To prove the given equation sinx + tanx = tanx (1 + cosx), you have simplified the left side (LS) of the equation to tanx (cosx + sinx).
To continue, we need to simplify the right side (RS) of the equation as well.
RS: tanx (1 + cosx)
Next, we can distribute tanx across the parentheses:
RS: tanx + tanx cosx
Now, let's compare LS and RS:
LS: tanx (cosx + sinx)
RS: tanx + tanx cosx
Both LS and RS have the term tanx in common. Now, we can factor out tanx from RS:
RS: tanx (1 + cosx)
Notice that the result on the right side of the equation is the same as LS.
Therefore, we have shown that sinx + tanx = tanx (1 + cosx) by simplifying both sides of the equation and showing that they are equal.