Posted by Rani on Tuesday, December 2, 2008 at 9:33pm.
i 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
the 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
PYTHAGOREAN 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.
oh, ok, I see.
thank you all very much. I think I get it now, so hopefully I can actually get them on my own. :)
wait, for the first one if sin^2+cos^2=1, how did you even cancel some out?
I'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.
okay, I'll try it out, thank you.
I'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.
You 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.)
but if I have identical denominators, why can't I just add the 1 to the tan^2X?
1+tan^2x/tan^2x
^ I split that fraction, making the 1/denom. a fraction, and adding it to the tan^2/denom fraction.
I haven't learned what cot and csc are. Can I still solve this with just tan, cos, and sin?
cot= 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
The 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!
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