# May i please have help verifying these equations. 1)cos(x+y)cos(x-y)=cos^22)1/(1-sin)=sec^2+tansec

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### 1 Answer

I think part of problem 1 got cut off.

`"cos" (x+y) = "cos"x "cos" y - "sin" x "sin" y`

`"cos" (x-y) = "cos"x "cos" y + "sin" x "sin" y`

So:`"cos" (x+y) "cos" (x-y) = ("cos"x "cos" y - "sin" x "sin"y ) ( "cos"x "cos" y + "sin" x "sin" y) `

` = "cos"^2 x "cos" ^2 y - "sin"^2 x "sin" ^2 y`

From here there are a couple things you could do, depending on what you're actually aiming for on the other side. Here's one option:

`"cos" ^2 y = 1 - "sin" ^2 y` , so

`"cos"^2 x "cos" ^2 y - "sin"^2 x "sin" ^2 y= "cos"^2 x (1 - "sin" ^2 y )- "sin"^2 x "sin" ^2 y `

`="cos"^2 x - "cos"^2 x "sin"^2 y - "sin"^2 x "sin"^2 y`

`="cos"^2 x - ("cos"^2 x + "sin" ^2 x)("sin"^2y)`

`="cos"^2 x - "sin"^2 y`

Similarly, you could do:

`"cos"^2 x "cos" ^2 y - "sin"^2 x "sin" ^2 y=(1-"sin"^2 x)"cos"^2 y - "sin"^2 x "sin"^2 y `

`="cos"^2 y - "sin"^2x "cos"^2y-"sin"^2x "sin"^2y`

`="cos"^2 y - "sin"^2 x("cos"^2 y + "sin"^2 x)`

`="cos"^2 y - "sin"^2 x`

Hopefully one of these is the identity that got cut off.

For problem 2:

First use the fact that `"sec" x = (1)/("cos" x)` and `"tan" x = ("sin" x)/("cos" x)` :

`"sec"^2 x + "tan" x "sec" x = (1)/("cos"^2 x)+(("sin" x)/("cos" x))((1)/("cos" x))=(1+"sin" x)/("cos"^2 x)`

Then use the identity `"sin"^2 x + "cos" ^2 x = 1` , so `"cos" ^2 x = 1 - "sin"^2 x`

`(1+"sin" x)/("cos"^2 x)=(1+"sin" x)/(1-"sin"^2 x)`

Factor: `1-"sin"^2 x = (1+"sin"x)(1-"sin"x)`

`(1+"sin" x)/(1-"sin"^2 x)=(1+"sin"x)/((1+"sin" x)(1-"sin"x))=(1)/(1-"sin"x)`

So: `"sec"^2 x+"tan"x "sec" x = (1)/(1-"sin" x)`