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Posted by **Jasmine** on Sunday, January 3, 2010 at 12:39pm.

Tan^2x-tan^2y=sec^2x-sec^2y

and, how do you factor and simplify,

cscx(sin^2x+cos^2xtanx)/sinx+cosx

- Trigonometry -
**Reiny**, Sunday, January 3, 2010 at 1:20pmOne of the basic identities is

tan^2 A + 1 = sec^2 A which gives Tan^2 A = sec^2 A - 1

so

Left side

= tan^2x-tan^2y

= sec^2x - 1 -(sec^2y - 1)

= sec^2x - sec^2y

= Right Side

- Trigonometry -
**Reiny**, Sunday, January 3, 2010 at 1:25pmFor the second, I will assume you meant

cscx(sin^2x+cos^2xtanx)/(sinx+cosx)

= 1/sinx(sin^2x + cos^2xsinx/cosx)/(sinx + cosx)

= 1/sinx(sin^2x + sinxcosx)/(sinx+cosx)

= (sinx + cosx)/sinx + cosx)

= 1

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