Question

Verify the identity. 1 - sin x / cos x = cos x / 1+ sin x

Answer 1

\[\frac{1-\sin x}{\cos x}=\cdots\\
\frac{1-\sin x}{\cos x}\cdot\frac{1+\sin x}{1+\sin x}=\cdots\\
\frac{1-\sin^2x}{\cos x(1+\sin x)}=\cdots\\
\frac{\cos^2x}{\cos x(1+\sin x)}=\cdots\]

Answer 2

Is that it ?

Answer 3

You have a factor of \(\cos x\) in both numerator and denominator, so you can remove those, leaving you with the desired right side of the equation.

Answer 4

How would I do that

Answer 5

Do you understand why \(\dfrac{x}{x}=1\)? You would apply this reasoning to say that
\[\frac{\cos^2x}{\cos x(1+\sin x)}=\cdots\\
\frac{\cos x}{\cos x}\cdot\frac{\cos x}{1+\sin x}=\cdots\\
1\cdot\frac{\cos x}{1+\sin x}=\cdots\\
\frac{\cos x}{1+\sin x}=\cdots\]

Answer 6

That would equal 1 + cotx right ? If That does then it would be 1 +cot x = cscx

Answer 7

No. There are no cotangents involved here.

Also, \(1+\cot x\not=\csc x\). You're thinking of \(1+\cot^2x=\csc^2x\).

Answer 8

I dont follow I am sorry can u just tell me the answer but explain it step by step please ?

Answer 9

You start off with
\[\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}\]
Notice the factor of \(1+\sin x\) in the denominator on the right side. That should be a clue as to what to do for the first step. How do you get \(1+\sin x\) in the denominator on the left side? Multiply the \(\cos x\) on the left by \(1+\sin x\). However, whatever you do the denominator must be done to the numerator to maintain equality. So, you multiply the left side by \(\dfrac{1+\sin x}{1+\sin x}\):
\[\frac{1+\sin x}{1+\sin x}\cdot\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}\]
This doesn't change the value of the left side, since (anything)/(same anything) = 1, and multiplying anything by 1 does not change the value. Make sense so far?

Answer 10

Yeah

Answer 11

Now we distribute:
\[\frac{1+\sin x}{1+\sin x}\cdot\frac{1-\sin x}{\cos x}=\frac{\cos x}{1+\sin x}\]
becomes
\[\frac{(1+\sin x)(1-\sin x)}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\\
\frac{1-\sin^2x}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\]
The numerator is a difference of squares. \((a+b)(a-b)=a^2-b^2.\)

Next, we use the following Pythagorean identity, \(\sin^2x+\cos^2x=1\). Rearranging a bit and solving for \(1-\sin^2x\), we have \(1-\sin^2x=\cos^2x\). So, we substitute:
\[\frac{\color{red}{1-\sin^2x}}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\]
becomes
\[\frac{\color{red}{\cos^2x}}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\]
Are you still with me?

Answer 12

\[\frac{\cos^2x}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\]
is the same as
\[\frac{\cos x\cos x}{\cos x(1+\sin x)}=\frac{\cos x}{1+\sin x}\]
or
\[\frac{\cos x}{\cos x}\cdot\frac{\cos x}{1+\sin x}=\frac{\cos x}{1+\sin x}\]
And since \(\dfrac{\cos x}{\cos x}=1\), we are left with
\[\frac{\cos x}{1+\sin x}=\frac{\cos x}{1+\sin x}\]
Thus the identity (equality) is verified, and we're done.

Answer 13

Ok thanks so much I think I have one more question if you could still help me that would be great if not thats fine

Answer 14

Sure, would you mind posting it as another question? In case I can't help, someone else might be able to.

Answer 15

Yeah I will I am just writing down everything for the problem

Answer 16

Okay, and you're welcome!

Answer 17

Thanks again