I'd greatly appreciate is somebody proved the following true.
1. 1-sinx/1+sinx = (secx-tanx)^2
2. sinx+1/cosx+cotx = tanx
3. sin^2x-tan^2x/1-sec^2x = sin^2x
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For second and third problems, how does the second step turn into the third step.
Originally Posted by fam600 2. sinx+1/cosx+cotx = tanx For the love of God, PLEASE use parenthesis! I wasted more than 10 minutes struggling with this before I understood what you meant. Your question should be typed as
(sinx+1)/(cosx+cotx) = tanx
If you don't understand the confusion, then please review order of operations.
Originally Posted by fam600 3. sin^2x-tan^2x/1-sec^2x = sin^2x Let me guess:
(sin^2x-tan^2x)/(1-sec^2x) = sin^2x
Different approaches . . .
Multiply top and bottom by . . . . . .
We have: .
We have: .
Thanks alot everyone. Sorobon, those different approaches especially helped. Its been a year since my last algebra course and so i've forgotten some of it, thats why i've been having such a hard time with these.
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