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anonymous
 4 years ago
Prove that (1 + secx)(1  cosx) = sinxtanx??
anonymous
 4 years ago
Prove that (1 + secx)(1  cosx) = sinxtanx??

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0write everything in terms of sin and cos to see things better

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 1  cosx + secx  cosxsecx = secx  cosx. Now what??

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0i was thinking more of: \[1cos+seccos.sec=sin.tan\] \[1cos+seccos/cos=sin.sin/cos\] \[1cos+sec1=sin^2/cos\] \[cos+1/cos=sin^2/cos\] for starters

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but we don't have to do that. we have to work from one side, and expand that to get the other side. It has to be proven.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0this is working one side; I havent changed the left at all except to see it better

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0and by left i mean right lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[cos+1/cos=sin^2/cos\] \[(cos^2+1)/cos=sin^2/cos\] \[(1cos^2)/cos=sin^2/cos\] \[sin^2/cos=sin^2/cos\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh silly me it is a product!! i thought it was a fraction...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but we are not supposed to even supposed to do ANYTHING with the other side. It's like we're not supposed to know the other side.

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0your requirements are absurd

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0your not spose to know tha tan = sin/cos? your not spose to know that sin.sin = sin^2? your not spose to know what; any trig relations and identities? but somehow transform them into each other?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not that, yaar. Imagine they give you a question, saying expand ....., leaving it terms of sin/cos/tan. SO we just do that. we expand. but how??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(1+\frac{1}{a})(1a)=\frac{1}{a}a=\frac{1a^2}{a}\] now replace a by cosine and you should get it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0since the right side isnt changed; just revert it back into its origanl form and and leave it there; but the process is still the same ...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is much easier to do the algebra without the sines and cosines. then put them in at the end.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but how does that give us sinxtanx??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but what amistre said is important. it in not a matter of leaving one side untouched. if you can change one side, you can change it back. if you have to hand it in in a certain form, fine, but if you can alter one side you can presumably go backwards

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0what? how does sin.sin/cos = sin tan?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0by definition of what sin and tan ARE lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok we did the algebra part, now comes the trig part. replace a by cosine and get \[\frac{1\cos^2(x)}{\cos(x)}\] now come the only trig step, writing \[1\cos^2(x)=\sin^2(x)\] that is the one step that is trig

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0can know that 1cos^2 = sin^2 sat; its forbidden knowledge ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I GOT IT. I love you guys :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so you get \[\frac{1\cos^2(x)}{\cos(x)}=\frac{\sin^2(x)}{\cos(x)}=\frac{\sin(x)}{\cos(x)}\times \sin(x)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okayy, another question. given that tanx = p, find an expression, in terms of p, for cosec^2 x. GO.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i see what they want

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so is it 1 + 1/p^2 ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i don't think so let me draw a triangle. you should draw one too

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327848673010:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there is my triangle, the triangle where tan(x) = p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now you need the hypotenuse which you find by pythagoras

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1327848731602:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hypotenuse is \[\sqrt{p^2+1}\] you are looking for \[\csc(x)\] and that is \[\frac{\sqrt{p^2+1}}{p}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nonononon. See, cosec^2x = 1 + cot^2x = 1 + 1/tan^2x = 1 + 1/p^2, right??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh sorry you want cosecant squared, so it is the square of what i wrote, which is \[\frac{p^2+1}{p^2}=1+\frac{1}{p^2}\] yes you have it!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i forgot about the square, sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hahahahah, studd. Since you love me, you have to do me a favour. You have to join my group.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my openstudy group. I created it B) It's called, wait for it, HerpDerp.
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