Trig help?

[Ihatemath]

I'm having some trouble with trig identities and whatnot. :/

Mainly questions like these:

Simplify: secx + tanx / 1/1-sinx.

I know the identities and I know that I should simplify it as: 1/cosx + sinx/cosx / 1/1-sinx, but then what do I do? Can I bring 1/cosx and sinx/cosx to a common factor; 1 + sinx/cosx? I find it hard to wrap my mind around fractions over fractions because it looks so weird and I never know what to do.

 

[galactus]

Please use proper grouping symbols(parentheses).

Is this what you mean?:

\(\displaystyle \frac{sec(x)+tan(x)}{\frac{1}{1-sin(x)}}\)

\(\displaystyle \frac{sec(x)}{\frac{1}{1-sin(x)}}+\frac{tan(x)}{\frac{1}{1-sin(x)}}\)

\(\displaystyle \frac{sec(x)}{\frac{1}{1-sin(x)}}=\frac{\frac{1}{cos(x)}}{\frac{1}{1-sin(x)}}=\frac{1}{cos(x)}\cdot\frac{1-sin(x)}{1}=\frac{1-sin(x)}{cos(x)}\)

\(\displaystyle \frac{tan(x)}{\frac{1}{1-sin(x)}}=\frac{\frac{sin(x)}{cos(x)}}{\frac{1}{1-sin(x)}}=\frac{sin(x)}{cos(x)}\cdot\frac{1-sin(x)}{1}=\frac{sin(x)-sin^{2}(x)}{cos(x)}\)

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

 

[Loren]

secx + tanx / 1/1-sinx means \(\displaystyle \sec x + \frac{\frac{\tan x}{1}}{1}-\sin x\) or \(\displaystyle \sec x + \frac{\tan x}{\frac{1}{1}}-\sin x\). You need to use grouping symbols correctly.

 

[Deleted member 4993]

I'm having some trouble with trig identities and whatnot. :/

Mainly questions like these:

Simplify: secx + tanx / 1/1-sinx.

if you meant:

\(\displaystyle \frac{\sec(x) \, + \, \tan(x)}{\frac{1}{1-sinx}}}\)

then you should have written:

[sec(x) + tan(x)]/[1/{1-sin(x)}]

I know the identities and I know that I should simplify it as: 1/cosx + sinx/cosx / 1/1-sinx, but then what do I do? Can I bring 1/cosx and sinx/cosx to a common factor; 1 + sinx/cosx? I find it hard to wrap my mind around fractions over fractions because it looks so weird and I never know what to do.

 

[soroban]

Hello, Ihatemath!

\(\displaystyle \text{Simplify: }\;\frac{\sec x + \tan x}{\dfrac{1}{1-\sin x}}\)

\(\displaystyle \text{First, get rid of the division: }\;(\sec x + \tan x)(1 - \sin x)\)


\(\displaystyle \text{Then we have: }\;\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right)(1 - \sin x)\)


\(\displaystyle \text{Multiply: }\;\frac{1}{\cos x} - \frac{\sin x}{\cos x} + \frac{\sin x}{\cos x} - \frac{\sin^2\!x}{\cos x}\)


\(\displaystyle \text{Simplify: }\;\frac{1}{\cos x} - \frac{\sin^2\!x}{\cos x} \;\;=\;\;\frac{\overbrace{1-\sin^2\!x}^{\text{This is }\cos^2\!x}}{\cos x} \;\;=\;\;\frac{\cos^2\!x}{\cos x} \;\;=\;\;\cos x\)