S somanyquestions Nov 2009 42 0 Mar 21, 2010 #1 verify that (cos x +1)/(tan^2 x) = (cos x)/(sec x -1) i'm having way too much trouble with these problems.

verify that (cos x +1)/(tan^2 x) = (cos x)/(sec x -1) i'm having way too much trouble with these problems.

pickslides MHF Helper Sep 2008 5,237 1,625 Melbourne Mar 21, 2010 #2 somanyquestions said: i'm having way too much trouble with these problems. Click to expand... always start with the left hand side and manipulate to find the right hand side. Here's a start... \(\displaystyle \frac{\cos x +1}{\tan^2 x}= \frac{\cos x }{\tan^2 x}+\frac{1}{\tan^2 x}= \dots\)

somanyquestions said: i'm having way too much trouble with these problems. Click to expand... always start with the left hand side and manipulate to find the right hand side. Here's a start... \(\displaystyle \frac{\cos x +1}{\tan^2 x}= \frac{\cos x }{\tan^2 x}+\frac{1}{\tan^2 x}= \dots\)

S somanyquestions Nov 2009 42 0 Mar 21, 2010 #3 pickslides said: always start with the left hand side and manipulate to find the right hand side. Here's a start... \(\displaystyle \frac{\cos x +1}{\tan^2 x}= \frac{\cos x }{\tan^2 x}+\frac{1}{\tan^2 x}= \dots\) Click to expand... well i know we can manipulate the left side in a few ways but it requires creativity and i can't do it.

pickslides said: always start with the left hand side and manipulate to find the right hand side. Here's a start... \(\displaystyle \frac{\cos x +1}{\tan^2 x}= \frac{\cos x }{\tan^2 x}+\frac{1}{\tan^2 x}= \dots\) Click to expand... well i know we can manipulate the left side in a few ways but it requires creativity and i can't do it.

S somanyquestions Nov 2009 42 0 Mar 21, 2010 #5 well i hate it because i can never do it i wish my teacher would go over more problems with us.