Prove that (1 + secx)(1 - cosx) = sinxtanx??

- anonymous

Prove that (1 + secx)(1 - cosx) = sinxtanx??

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- schrodinger

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- amistre64

write everything in terms of sin and cos to see things better

- anonymous

so 1 - cosx + secx - cosxsecx = secx - cosx. Now what??

- anonymous

sec=cos/sin

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- anonymous

no, sec is 1/cos

- anonymous

hhaha

- amistre64

i was thinking more of:
\[1-cos+sec-cos.sec=sin.tan\]
\[1-cos+sec-cos/cos=sin.sin/cos\]
\[1-cos+sec-1=sin^2/cos\]
\[-cos+1/cos=sin^2/cos\]
for starters

- anonymous

but 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

this is working one side; I havent changed the left at all except to see it better

- amistre64

and by left i mean right lol

- amistre64

\[-cos+1/cos=sin^2/cos\]
\[(-cos^2+1)/cos=sin^2/cos\]
\[(1-cos^2)/cos=sin^2/cos\]
\[sin^2/cos=sin^2/cos\]

- anonymous

oh silly me it is a product!! i thought it was a fraction...

- anonymous

but 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.

- anonymous

^that's right :P

- amistre64

your requirements are absurd

- amistre64

your 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

not 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

\[(1+\frac{1}{a})(1-a)=\frac{1}{a}-a=\frac{1-a^2}{a}\] now replace a by cosine and you should get it

- amistre64

since 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

it is much easier to do the algebra without the sines and cosines. then put them in at the end.

- anonymous

but how does that give us sinxtanx??

- anonymous

but 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

what? how does sin.sin/cos = sin tan?

- amistre64

by definition of what sin and tan ARE lol

- anonymous

ok 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

can know that 1-cos^2 = sin^2 sat; its forbidden knowledge ;)

- anonymous

I GOT IT.
I love you guys :D

- anonymous

so you get
\[\frac{1-\cos^2(x)}{\cos(x)}=\frac{\sin^2(x)}{\cos(x)}=\frac{\sin(x)}{\cos(x)}\times \sin(x)\]

- amistre64

:) good luck

- anonymous

and we love you!

- anonymous

okayy, another question.
given that tanx = p, find an expression, in terms of p, for cosec^2 x.
GO.

- amistre64

gone!! ;)

- anonymous

??

- anonymous

oh i see what they want

- anonymous

so is it 1 + 1/p^2 ??

- anonymous

i don't think so let me draw a triangle. you should draw one too

- anonymous

|dw:1327848673010:dw|

- anonymous

there is my triangle, the triangle where tan(x) = p

- anonymous

now you need the hypotenuse which you find by pythagoras

- anonymous

|dw:1327848731602:dw|

- anonymous

hypotenuse is
\[\sqrt{p^2+1}\] you are looking for
\[\csc(x)\] and that is
\[\frac{\sqrt{p^2+1}}{p}\]

- anonymous

nonononon. See, cosec^2x = 1 + cot^2x = 1 + 1/tan^2x = 1 + 1/p^2, right??

- anonymous

oh 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

i forgot about the square, sorry

- anonymous

hahahahah, studd. Since you love me, you have to do me a favour. You have to join my group.

- anonymous

group of what?

- anonymous

my openstudy group. I created it B) It's called, wait for it, Herp-Derp.

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