Prove: 1+sinx/1-sinx=csc^2x+2cscx+1
Please Help! It's really important
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Just remember that \(\mathbb {cscx=\frac{1}{sinx}}\) and use \((a-b)(a+b)=a^2-b^2\)
is the proof solvable because iv'e tried everything?
if you mean (1+sinx)/(1-sinx)=csc^2x+2cscx+1 this is not an identity. It is true for specific values of x i.e. you can solve for the x's that make it true.
how could i solve for x
Up above you said Prove: Is this supposed to be an identity that they want you to prove ? (though it's not an identity, this may be a question with a "typo" and they meant to give an identity)
Or do they want you to "solve for x" ?
It's supposed to be a trig identity that he wants us to prove
I have been trying for hours to prove it
I tried and found out that this is not an identity. @phi is right
If the right side is \(sec^2x+2secx+1\), then it will be an identity. Is that right, @phi?
no, I don't think its sec^2 + 2sec + 1 (but if you can show it works, I'm all ears) I'm trying to figure out what they might have meant, but so far, no luck.
I will try
First, is it really \[ \frac{1+\sin x}{1-\sin x}= \csc^2 x+2\csc x+1\]?
It says, PROVE: you can manipulate the left side to make it 'look' like the right side. Do not manipulate the right side.
yes that is what the question is
what do you get if you let x=30 degrees for both sides ?
Yeah, @phi is right. Neither way is an identity
sin(30)= 0.5 (1+0.5)/(1-0.5) = 3 csc(x)= 1/sin(x) so csc(30) = 2 csc^2(30)= 4 the right side is csc^2(30) + 2 csc(30) + 1 = 4 +2 +1 = 7 3 does not equal 7 that proves it's not an identity
Okay! Thank you so much for your help!
**the right side is csc^2(30) + 2 csc(30) + 1 = 4 +2*2 +1 = 9 3 does not equal 9 that proves it's not an identity
It's just weird that he would ask that question!
so it might be a "mistake" or he wanted you to show it's not true.
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