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# C3 Trig Question watch

1. How do I do...

(1 - cosx)(1+secx) = sinxtanx

Should I expand the left hand side? This is what I have done...but I still can't get it so that it's like the RHS where sinx x sinx/cosx

Thank you
2. Where did you get the equation from? It's not a trig identity that holds for all x.
3. (Original post by ttoby)
Where did you get the equation from? It's not a trig identity that holds for all x.
Whoops, mistake. It's + secx not - secx.

And from the edexcel c3 textbook
4. Expand the LHS, should come out pretty easily.
5. This is how I did it...

sin(x)tan(x)= sin^2(x)/cos(x)= (1-cos^2(x))/cos(x)= 1/cosx - cosx = secx - cosx = LHS

Hopefully that helps.
6. ok doing it nw
7. (Original post by insparato)
Expand the LHS, should come out pretty easily.
Got it!

Solved.
8. just start on the left hand side expand the brackets then simplify.
9. When i did it i got down to secx - cosx but whats the last and final step to get it into sinxtanx?
10. Combine that.
11. expand the brackets (LHS):
=1+secx-cosx-(cosx)(secx)
=1+(1/cosx)-cosx-cosx(1/cosx)
=1+(1/cosx)-cosx-1
=(1/cosx)-cosx
=(1/cosx)-(cosx/1)
=(1-cos^2x)/cosx <--cross multiply to get same denominator
=sin^2x/cosx <--use of identity sin^2x +cos^2x =1
=tanxsinx

gd luck wid revision + exam on wednesday

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