J justseel New member Joined Apr 8, 2014 Messages 2 Apr 8, 2014 #1 My math teacher ask me to establish the identity in two ways from right side to the left side and from left ti right sec=sin/cos + cos/1+sin
My math teacher ask me to establish the identity in two ways from right side to the left side and from left ti right sec=sin/cos + cos/1+sin
D Deleted member 4993 Guest Apr 8, 2014 #2 justseel said: My math teacher ask me to establish the identity in two ways from right side to the left side and from left ti right sec=sin/cos + cos/1+sin Click to expand... What have you tried? What are your thoughts? Please share your work with us ... If you are stuck at the beginning tell us and we'll start with the definitions. You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL: http://www.freemathhelp.com/forum/th...Before-Posting
justseel said: My math teacher ask me to establish the identity in two ways from right side to the left side and from left ti right sec=sin/cos + cos/1+sin Click to expand... What have you tried? What are your thoughts? Please share your work with us ... If you are stuck at the beginning tell us and we'll start with the definitions. You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL: http://www.freemathhelp.com/forum/th...Before-Posting
J justseel New member Joined Apr 8, 2014 Messages 2 Apr 9, 2014 #3 yes I,m stuck at the beginning i tired to do this but i didnt know how end it 1/cosx = sinx+1/cosxsinxcosx
yes I,m stuck at the beginning i tired to do this but i didnt know how end it 1/cosx = sinx+1/cosxsinxcosx
R richardt Junior Member Joined Aug 27, 2013 Messages 56 Apr 13, 2014 #4 Greetings: A hint: Try multiplying numerator and denominator of cosx/(1+sinx) by the conjugate of 1 + sinx. Rich Last edited: Apr 13, 2014
Greetings: A hint: Try multiplying numerator and denominator of cosx/(1+sinx) by the conjugate of 1 + sinx. Rich
D Deleted member 4993 Guest Apr 13, 2014 #5 Enough time has gone by: Right to left \(\displaystyle \dfrac{sin(\theta)}{cos(\theta)} + \dfrac{cos(\theta)}{1+sin(\theta)}\) \(\displaystyle = \ \dfrac{sin(\theta)*[1+sin(\theta)] + cos^2(\theta)}{cos(\theta)*[1+sin(\theta)]} \) \(\displaystyle = \ \dfrac{1+sin(\theta)}{cos(\theta)*[1+sin(\theta)]} \) \(\displaystyle = \ \dfrac{1}{cos(\theta)} \) \(\displaystyle = \ sec(\theta) \) Now just go back-wards.... for left-to-right
Enough time has gone by: Right to left \(\displaystyle \dfrac{sin(\theta)}{cos(\theta)} + \dfrac{cos(\theta)}{1+sin(\theta)}\) \(\displaystyle = \ \dfrac{sin(\theta)*[1+sin(\theta)] + cos^2(\theta)}{cos(\theta)*[1+sin(\theta)]} \) \(\displaystyle = \ \dfrac{1+sin(\theta)}{cos(\theta)*[1+sin(\theta)]} \) \(\displaystyle = \ \dfrac{1}{cos(\theta)} \) \(\displaystyle = \ sec(\theta) \) Now just go back-wards.... for left-to-right