Basic Trigonometry Identities: solve sec^2x-1/1-csc^2x

[andie]

The equation I need to solve is sec^2x-1/1-csc^2x
I was told I had to substitute top and bottom with a sort of identity.
I was thinking using the Pythagorean identities where cos^2x=1-sin^2x and sin^2x=1-cos^2x, since secx=1/cosx and cscx=1/sinx.
I'm not sure if that's right though and even if it is, I don't know how to substitute it in.

 

[Loren]

Re: Basic Trigonometry Identities

sec^2x-1/1-csc^2x means \(\displaystyle \sec^2x - \frac{1}{1}-\csc^2x\). I know you don't mean that so you need to use grouping symbols to clarify. Furthermore, that isn't an equation.

 

[Deleted member 4993]

The equation I need to solve is sec^2x-1/1-csc^2x

If you meant:

\(\displaystyle \frac{\sec^2(x) \, - \, 1}{\csc^2(x) \, - \, 1} \,\)

then in ASCII format you should wrtite it as

[sec^2(x) - 1]/[csc^2(x) - 1]

The parentheses eliminate all the confusion.

 

[soroban]

Hello, andie!

Warning: "Solve the equation" does not apply to ALL math problems!

\(\displaystyle \text{Simplify: }\:\frac{\sec^2\!x-1}{1-\csc^2\!x}\)

Are you familiar with the other Pythagorean identities?

. . \(\displaystyle \sec^2\!x \:=\:\tan^2\!x + 1\)
. . \(\displaystyle \csc^2\!x \:=\:\cot^2\!x + 1\)


\(\displaystyle \text{We have: }\:\frac{\sec^2\!x-1}{1-\csc^2\!x} \;=\;\frac{\sec^2\!x-1}{-(\csc^2\!x-1)} \;=\;\frac{\tan^2\!x}{-\cot^2\!x} \;=\;\frac{\tan^2\!x}{-\frac{1}{\tan^2x}} \;=\;-\tan^4\!x\)