i need more help!!! (sorry)

[yasaminG]

ok so im just beginning to get the hang of these problems, but i get stuck after the 2nd step :


@=theta

sec@-csc@ / sec@+csc@ = tan@-1/tan@+1

we have to prove that the left side is equal to the right.
thank you!!

 

[stapel]

Do you mean what you have posted?

. . . . .$\displaystyle \large{\sec{(\theta)} - \frac{\csc{(\theta)}}{\sec{(\theta)}} + \csc{(\theta)} = \tan{(\theta)} - \frac{1}{\tan{(\theta)}} + 1}$

...or something more along the lines of:

. . . . .\(\displaystyle \large{\frac{\sec{(\theta)} - \csc{(\theta)}}{\sec{(\theta)} + \csc{(\theta)}} = \frac{\tan{(\theta)} - 1}{\tan{(\theta)} + 1}\)

...or something else?

Please reply with clarifications, showing what you have tried thus far. Thank you.

Eliz.

 

[soroban]

Hello, yasaminG!

$\displaystyle \frac{\sec\theta\,-\,\csc\theta}{\sec\theta\,+\,\csc\theta}\:=\:\frac{\tan\theta\,-\,1}{\tan\theta\,+\,1}$

The left side is: .\(\displaystyle \L\frac{\frac{1}{\cos\theta}\,-\,\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}\,+\,\frac{1}{\sin\theta}}\)

Multiply top and bottom by \(\displaystyle \sin\theta:\;\;\L\frac{\frac{\sin\theta}{\cos\theta}\,-\,1}{\frac{\sin\theta}{\cos\theta}\,+\,1}\;=\;\frac{\tan\theta\,-\,1}{\tan\theta\,+\,1}\)

 

[yasaminG]

Hello, yasaminG!

$\displaystyle \frac{\sec\theta\,-\,\csc\theta}{\sec\theta\,+\,\csc\theta}\:=\:\frac{\tan\theta\,-\,1}{\tan\theta\,+\,1}$

The left side is: .\(\displaystyle \L\frac{\frac{1}{\cos\theta}\,-\,\frac{1}{\sin\theta}}{\frac{1}{\cos\theta}\,+\,\frac{1}{\sin\theta}}\)

Multiply top and bottom by \(\displaystyle \sin\theta:\;\;\L\frac{\frac{\sin\theta}{\cos\theta}\,-\,1}{\frac{\sin\theta}{\cos\theta}\,+\,1}\;=\;\frac{\tan\theta\,-\,1}{\tan\theta\,+\,1}\)

thank you, your explanation makes them so easy!