Question

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

Answer 1

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

Answer 2

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

Answer 3

sec=cos/sin

Answer 4

no, sec is 1/cos

Answer 5

hhaha

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

and by left i mean right lol

Answer 10

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

Answer 11

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

Answer 12

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.

Answer 13

^that's right :P

Answer 14

your requirements are absurd

Answer 15

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?

Answer 16

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??

Answer 17

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

Answer 18

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 ...

Answer 19

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

Answer 20

but how does that give us sinxtanx??

Answer 21

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

Answer 22

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

Answer 23

by definition of what sin and tan ARE lol

Answer 24

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

Answer 25

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

Answer 26

I GOT IT.

I love you guys :D

Answer 27

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

Answer 28

:) good luck

Answer 29

and we love you!

Answer 30

okayy, another question.

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

GO.

Answer 31

gone!! ;)

Answer 32

??

Answer 33

oh i see what they want

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

now you need the hypotenuse which you find by pythagoras