Not Sure if I did this correctly with Trig Identities

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• October 29th 2012, 05:39 PM
kickyou77
Not Sure if I did this correctly with Trig Identities
Algebra seems like a blur, so I need a bit of help with this problem.(~)==theta
The original problem is this, and you are supposed to prove that it is true

cos(~)
------- = Sec(~)+tan(~)
1-sin(~)

my work:

1/sec * 1/1-sin(~)

1/sec(1-sin(~)

1/sec(~)-sec(~)*sin(~)

----This is where I got confused(I couldn't remember if you could use the inverse to make this statement true)------

(1/sec(~)-tan(~))^-1=Sec(~)+tan(~)

Could I get a yes or no if I did it right? Thanks
• October 29th 2012, 05:52 PM
skeeter
Re: Not Sure if I did this correctly with Trig Identities
$\frac{\cos{t}}{1 - \sin{t}} \cdot \frac{1+\sin{t}}{1+\sin{t}} =$

$\frac{\cos{t}(1+\sin{t})}{1-\sin^2{t}} =$

$\frac{\cos{t}(1+\sin{t})}{\cos^2{t}} =$

$\frac{1+\sin{t}}{\cos{t}} =$

$\frac{1}{\cos{t}} + \frac{\sin{t}}{\cos{t}} =$

$\sec{t} + \tan{t}$
• October 29th 2012, 05:57 PM
Prove It
Re: Not Sure if I did this correctly with Trig Identities
Quote:

Originally Posted by kickyou77
Algebra seems like a blur, so I need a bit of help with this problem.(~)==theta
The original problem is this, and you are supposed to prove that it is true

cos(~)
------- = Sec(~)+tan(~)
1-sin(~)

my work:

1/sec * 1/1-sin(~)

1/sec(1-sin(~)

1/sec(~)-sec(~)*sin(~)

----This is where I got confused(I couldn't remember if you could use the inverse to make this statement true)------

(1/sec(~)-tan(~))^-1=Sec(~)+tan(~)

Could I get a yes or no if I did it right? Thanks

\displaystyle \begin{align*} \frac{\cos{\theta}}{1 - \sin{\theta}} &\equiv \frac{\cos{\theta} \left( 1 + \sin{\theta} \right)}{\left( 1 - \sin{\theta} \right) \left( 1 + \sin{\theta} \right)} \\ &\equiv \frac{\cos{\theta}\left( 1 + \sin{\theta} \right)}{1 - \sin^2{\theta}} \\ &\equiv \frac{\cos{\theta} \left( 1 + \sin{\theta} \right)}{\cos^2{\theta}} \\ &= \frac{1 + \sin{\theta}}{\cos{\theta}} \\ &\equiv \frac{1}{\cos{\theta}} + \frac{\sin{\theta}}{\cos{\theta}} \\ &\equiv \sec{\theta} + \tan{\theta} \end{align*}
• October 29th 2012, 06:35 PM
Soroban
Re: Not Sure if I did this correctly with Trig Identities
Hello, kickyou77!

Quote:

$\text{Prove: }\:\frac{\cos\theta}{1-\sin\theta} \:=\:\sec\theta + \tan\theta$

You can also work from the right side . . .

$\sec\theta + \tan\theta \;=\; \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \;=\;\frac{1 + \sin\theta}{\cos\theta}$

. . . . . . . . . $=\;\frac{1 + \sin\theta}{\cos\theta}\cdot {\color{blue}\frac{1-\sin\theta}{1-\sin\theta}} \;=\;\frac{1-\sin^2\!\theta}{\cos\theta(1 - \sin\theta)}$

. . . . . . . . . $=\;\frac{\cos^2\!\theta}{\cos\theta(1-\sin\theta)} \;=\; \frac{\cos\theta}{1-\sin\theta}$