Question

Can anyone help me prove these identities?

Answer 1

\[\frac{ 1 }{ \sec x \tan x} =\csc x - sinx\]

Answer 2

@jim_thompson5910

Answer 3

@mathstudent55

Answer 4

say \(\bf sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=?\)

Answer 5

1 - cos^2x

Answer 6

@jdoe0001

Answer 7

yeap....thus

Answer 8

would sinx = 1 - cosx then?

Answer 9

\(\bf \cfrac{ 1 }{ sec (x) tan (x)} =csc( x) - sin(x)
\\ \quad \\ \quad \\
\cfrac{1}{\frac{1}{cos(x)}\cdot \frac{sin(x)}{cos(x)}}\implies \cfrac{1}{\frac{sin(x)}{cos^2(x)}}\implies \cfrac{cos^2(x)}{sin(x)}
\\ \quad \\
{\color{brown}{ sin^2(\theta)+cos^2(\theta)=1\implies sin^2(\theta)=1-cos^2(\theta)}}
\\ \quad \\
\cfrac{cos^2(x)}{sin(x)}\implies \cfrac{{\color{brown}{ 1-cos^2(x)}}}{sin(x)}\implies \cfrac{1}{sin(x)}-\cfrac{cos^2(x)}{sin(x)}\implies ?\)

Answer 10

hmm holld the mayo.... need to fix that a bit =) got a couple of typos hehe

Answer 11

okie haha

Answer 12

\(\bf \cfrac{ 1 }{ sec (x) tan (x)} =csc( x) - sin(x)
\\ \quad \\ \quad \\
\cfrac{1}{\frac{1}{cos(x)}\cdot \frac{sin(x)}{cos(x)}}\implies \cfrac{1}{\frac{sin(x)}{cos^2(x)}}\implies \cfrac{cos^2(x)}{sin(x)}
\\ \quad \\
{\color{brown}{ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)}}
\\ \quad \\
\cfrac{cos^2(x)}{sin(x)}\implies \cfrac{{\color{brown}{ 1-sin^2(x)}}}{sin(x)}\implies \cfrac{1}{sin(x)}-\cfrac{\cancel{ sin^2(x) }}{\cancel{ sin(x) }}\implies ?\)

Answer 13

okay I get itt :) I just couldn't get past 1-sin^2x/sinx
thank you so much

Answer 14

yw