how to solvveee:
91-cosx) (1+Secx)=sinx tanx
2007-11-24 11:57:45 · 4 answers · asked by ? 1 in Science & Mathematics ➔ Mathematics
(1-cosx) (1+Secx)
(1 - cosx)(1 + 1/cosx)
1 + 1/cosx - cosx - 1
1/cosx - cosx
1/cosx - cos^2x/cosx
(1 - cos^2x)/cosx
sin^2x/cosx
sinx(sinx/cosx)
sinxtanx
2007-11-24 12:05:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I'm guessing it's not a nine but a ( so...
first distribute:
1 + secx - cosx - cosx*secx = sinxtanx
simplify:
1 + 1/cosx - cosx - 1 = sinx*sinx/cosx
simplify:
1/cosx - cosx = sin^2 x / cosx
same denominator:
1/cosx - cos^2x/cosx = sin^2x / cosx
combine numberators:
(1 - cos^2 x) / cosx = sin^2x / cosx
simplify the numerator (use the sin^2x + cos^2x = 1)
sin^2x / cosx = sin^2x / cosx
2007-11-24 12:06:56 · answer #2 · answered by mathsmile 2 · 0⤊ 0⤋
1-cosx+secx-1
=secx-cosx
=i/cos - cosx
=(1-cos^2x)/cosx
=sin^2x/cosx
=sinx sinx/cosx
=sinx tanx
= rhs
2007-11-24 12:03:42 · answer #3 · answered by norman 7 · 0⤊ 0⤋
(1 -- cosx)(1 + secx)
= (1 -- cosx)(1 + cosx)/cosx
= (1 -- cos^2x)/cosx
= sin^2x/cosx
= sinxsinx/cosx
= sinxtanx
2007-11-24 12:08:38 · answer #4 · answered by sv 7 · 0⤊ 0⤋