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August 3, 2015

August 3, 2015

Posted by **Rani** on Tuesday, December 2, 2008 at 9:33pm.

Prove each idenity.

1+1/tan^2x=1/sin^2x

1/cosx-cosx=sinxtanx

1/sin^2x+1/cos^2x=1/sin^2xcos^2x

1/1-cos^2x+/1+cosx=2/sin^2x

and

(1-cos^2x)(1+1/tan^2x)= 1

I haven't even gotten 'round to sny of the quedtions because the first one is just so hard. I'm not really sure I'm uderstanding how to use the quotient and pythagorean identities. I'm so confused. I can't make sense of it and I have tried so many different ways. I must have spent over an hour on he first problem and still I can't come up with an answer. I'm just really frustrated; could someone please help me. I'd really appreciate it.

- Math - help really needed -
**Reiny**, Tuesday, December 2, 2008 at 9:55pmi will do the first one

1+1/tan^2x=1/sin^2x

LS = 1 + 1/(sin^2x/cos^2x)

= 1 + cos^2x/sin^2x

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

= 1/sin^2x

= RS

I usually try to change all ratios to sines and cosines

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**Reiny**, Tuesday, December 2, 2008 at 9:58pmthe second is quite easy.

add the left side terms by taking a common denominator of cosx

you will get

(1-cos^2x)/cosx

= sin^2x/cosx

= sinx(sinx/cosx)

= sinxtanx

= RS

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**Bob**, Tuesday, December 2, 2008 at 9:59pmPYTHAGOREAN IDENTITIES:

sin^2 + cos^2 = 1 (a^2 + b^2 = c^2)

sec^2 - tan^2 = 1

csc^2 = cot^2 = 1 (the co-version of the above)

(1/cosx)-(cosx/1)=sinxtanx

To subtract those two, you need a common denominator, so multiply the (cosx/1) by (cosx/cosx). When you subtract them, the new numerator is an identity, so it can be re-written. When you simplify, you're left with sinxtanx.

The rest are very similar. I hope you the idea of how to solve these types of problems - you just rearrange the complicated side to get identities that can be re-written.

- Math - help really needed -
**Rani**, Tuesday, December 2, 2008 at 10:08pmoh, ok, I see.

thank you all very much. I think I get it now, so hopefully I can actually get them on my own. :)

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**Rani**, Tuesday, December 2, 2008 at 10:17pmwait, for the first one if sin^2+cos^2=1, how did you even cancel some out?

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**Bob**, Tuesday, December 2, 2008 at 10:28pmI'll give you a hint:

Try replacing the 1 with tan^2x/tan^2x. After add that to 1/tan^2x, replace the denominator (the tan^2x) with sin^2/cos^2.

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**Rani**, Tuesday, December 2, 2008 at 10:35pmokay, I'll try it out, thank you.

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**Rani**, Tuesday, December 2, 2008 at 10:40pmI'm really sorry. I don't want to sound stupid, but when I add tan^2x/tan^2x to 1/tan^2x I get 1+tan^2x/tan^2x and then the tans cancel out and I'm left with just one. Obviously, that's not correct, but I can't find my mistake.

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**Bob**, Tuesday, December 2, 2008 at 11:26pmYou actually get (1/tan^2x)+1, which is cot^2 + (csc^2x-cot^2x), which is csc^2x, which is 1/sin^2x. And that's your answer.

(You could only cancel out the tan^2 if it were (Some #)(tan^2x)/tan^2x, rather than (1)+(tan^2x)/tan^2x. Dividing is the inverse of multiplying, so dividing only cancels what is multiplied, not added.)

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**Rani**, Tuesday, December 2, 2008 at 11:28pmbut if I have identical denominators, why can't I just add the 1 to the tan^2X?

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**Bob**, Tuesday, December 2, 2008 at 11:33pm1+tan^2x/tan^2x

^ I split that fraction, making the 1/denom. a fraction, and adding it to the tan^2/denom fraction.

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**Rani**, Tuesday, December 2, 2008 at 11:37pmI haven't learned what cot and csc are. Can I still solve this with just tan, cos, and sin?

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**Anonymous1**, Wednesday, December 3, 2008 at 12:15amcot= 1/tan or inverse of tangent

csc= 1/sin or inverse of sine

(1-cos^2x)(1+1/tan^2x)= 1

(1-cos^2x)= sin x

(1+1/tan^2x)= cscx

So

(sin x)(1/sin x)=1

1=1

- Math - help really needed -
**Cianán**, Thursday, December 4, 2008 at 1:14pmThe first one is simpler if you do it this way:

Multiply both sides by [sin(x)]^2

then:

1+1/tan^2x=1/sin^2x

becomes

sin^2x + sin^2x/tan^2x = 1

since tan^2x = sin^2x/cos^2x the equation becomes:

sin^2x + sin^2x/(sin^2x/cos^2x) = 1

sin^2x + sin^2x*cos^2x/sin^2x = 1

the two sin^2x cancels and you are left with:

sin^2x + cos^2x = 1 which is true!

Hope that helps!