Question

Trigg identity

Answer 1

(secx + tanx)(1 - sinx / cosx) = 1
I am supposed to prove the identity. I got it down to:
(1 + sinx)/cosx = 1
Which is 1 + sinx = cosx ... which isn't true... where did I go wrong?

Answer 2

Is it \(\color{blue}{\displaystyle (\sec x +\tan x)\cdot \left(\frac{1-\sin x}{\cos x}\right) }\) ?

Answer 3

Yes!

Answer 4

1 - sinx = cosx, right?

Answer 5

Ok, you know the following:

1) tan(x) = sin(x) / cos(x)
2) sec(x) = 1 / cos(x)

So, you can factor 1/cos(x) out.

\(\color{blue}{\displaystyle (\sec x +\tan x)\cdot \left(\frac{1-\sin x}{\cos x}\right)=1 }\)

\(\color{blue}{\displaystyle \left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right) \cdot \left(\frac{1-\sin x}{\cos x}\right)=1 }\)

\(\color{blue}{\displaystyle \left(\frac{1+\sin x}{\cos x}\right) \cdot \left(\frac{1-\sin x}{\cos x}\right)=1 }\)

\(\color{blue}{\displaystyle \frac{(1+\sin x)(1-\sin x)}{\cos^2x}=1 }\)

Answer 6

this is all you need...

using algebra, you can easily simplify the top, and then re-write it using the basic trig identity you know. If you do it correctly, it should verify to be equivalent.

If you have questions, ask.

Answer 7

That looks similar to what I did... except I replaced the 1 - sinx with cosx. Was that where I went wrong?

Answer 8

Oh, because that is not a correct replacement to make.

Answer 9

Doesn't cosx + sinx = 1 ? :/

Answer 10

Oh, no ...

Answer 11

\(\sin^2x+\cos^2x=1\)

Answer 12

but, not just with sin and cos.

Answer 13

Oh shoot! That's where I went wrong... okay, thank you

Answer 14

Not a problem ....

then the difference of squares
\((1+\sin x)(1-\sin x):=1^2-(\sin x)^2=1-\sin^2x\)

and knowing the aforementioned property (\(\sin^2x+\cos^2x=1\)), you know that
\(1-\sin^2x=\cos^2x\)

Answer 15

Right, right :)

I saw your name and now I'm craving salmon.

Answer 16

You prefer fish over meat? (What .. ?! ;

Answer 17

Oooo salmon :o my fav

Answer 18

YESS Zep. Salmon <3 <3 <3