I NEED HELP!!! (sinx-1)(tanx+secx)= -cosx Should I substitute?
L Lenore New member Joined Feb 27, 2010 Messages 1 Feb 27, 2010 #1 I NEED HELP!!! (sinx-1)(tanx+secx)= -cosx Should I substitute?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Feb 27, 2010 #2 Expand out and use some identities. \(\displaystyle (sin(x)-1)(tan(x)+sec(x))\) \(\displaystyle sin(x)tan(x)+\underbrace{\frac{sin(x)}{cos(x)}}_{\text{tan(x)}}-tan(x)-sec(x)\) \(\displaystyle sin(x)tan(x)-sec(x)\) \(\displaystyle \frac{sin^{2}(x)}{cos(x)}-\frac{1}{cos(x)}\) \(\displaystyle \frac{sin^{2}(x)-1}{cos(x)}\) \(\displaystyle \frac{-cos^{2}(x)}{cos(x)}\) \(\displaystyle -cos(x)\)
Expand out and use some identities. \(\displaystyle (sin(x)-1)(tan(x)+sec(x))\) \(\displaystyle sin(x)tan(x)+\underbrace{\frac{sin(x)}{cos(x)}}_{\text{tan(x)}}-tan(x)-sec(x)\) \(\displaystyle sin(x)tan(x)-sec(x)\) \(\displaystyle \frac{sin^{2}(x)}{cos(x)}-\frac{1}{cos(x)}\) \(\displaystyle \frac{sin^{2}(x)-1}{cos(x)}\) \(\displaystyle \frac{-cos^{2}(x)}{cos(x)}\) \(\displaystyle -cos(x)\)
