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October 1, 2014

October 1, 2014

Posted by **Don** on Sunday, January 8, 2012 at 7:38pm.

2 columns

(tan + cot)^2 = sec^2 + csc^2

I'm having trouble breaking down the left side to = the right side..

Any help please

- Trig -
**Damon**, Sunday, January 8, 2012 at 7:49pmleft

(sin/cos + cos/sin)^2

sin^2/cos^2 + 2 + cos^2/sin^2

[sin^4 +2sin^2 cos^2+cos^4 ]/cos^2 sin^2

(sin^2+cos^2)^2/cos^2sin^2

1^2/sin^2cos^2

1/sin^2 cos^2

right

1/cos^2 + 1/sin^2

sin ^2/cos^2sin^2 + cos^2/cos^2 sin^2

1/cos^2 sin^2

- Trig -
**Don**, Sunday, January 8, 2012 at 8:21pmHi Damon .. apparently they want the right side to stay "as is" and for the left side to transform into exactly what the right side says .... sorry

- Trig -
**Damon**, Sunday, January 8, 2012 at 8:40pmI do not think so. That would be a very unusual thing for "them" to say :)

- Trig -
**Don**, Sunday, January 8, 2012 at 8:54pmThe question says: Set up a 2 column proof to show that each of the following equations is an identity. Transform the left side to become the right side.

(tan + cot)^2 = sec^2 + csc^2

- Trig -
**Steve**, Monday, January 9, 2012 at 5:10am(tan + cot)^2 = tan^2 + 1 + cot^2

= sec^2 - 1 + 2 + csc^2 - 1

= sec^2 + csc^2

- Trig - oops -
**Steve**, Monday, January 9, 2012 at 5:11amoops that's tan^2 + 2 + cot^2

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