Proving identities:
1) 1+ 1/tan^2x = 1/sin^2x
2) 2sin^2 x-1 = sin^2x - cos^2x
3) 1/cosx - cosx = sin x tan x
4) sin x + tan x =tan x (1+cos x)
5) 1/1-sin^2x= 1+tan^2 x
How in the world do I prove this...please help...
I appreciateyour time thankyou soo much!!
There is on one correct and foolproof way to prove identities.
There are some general rules you might follow
1. look for obvious relations , like sin^2 x + cos^2 x = 1
or 1 + tan^2 x = sec^2 x
-- make yourself a summary of these collected from your text or notebooks
2. usually, changing all ratios to sines and cosines often gives quick and easy results
3. start with the more complicated looking side, and try to work it towards the expression on the other side , or
4. work down one side until you can't seem to go any further, now switch to the other side and try to obtain that last expression
e.g. #4
sinx + tanx = tanx (1+cosx)
LS = sinx + sinx/cosx
= (sinxcosx + sinx)/cosx
= sinx(cosx + 1)/cosx
= (sinx/cosx)(1 + cosx)
= tanx (1 + cosx)
= RS
#2
2sin^2 x - 1 = sin^2x - cos^2x
RS = sin^2 x - (1 - sin^2 x)
= sin^2 x - 1 + sin^2 x
= 2sin^2 x - 1
= LS
well, that was an easy one
try the others by using similar methods.
come back if you get stuck.
My first line should have been:
There is NO one correct and foolproof way to prove identities.
OHHH!!! Ok thankyou soooo much I greatly appreciate your help and one compliment...Your way better than my teacher...once again thankyou sooo much I think I get the jist of it now:)
To prove identities, we simplify one side of the equation until it is equivalent to the other side. Here's how you can prove each of the given identities:
1) Start with the left side of the equation:
LHS: 1 + 1/tan^2x
Recall that tan^2x = sin^2x / cos^2x, so replace tan^2x with sin^2x / cos^2x.
LHS: 1 + 1 / (sin^2x / cos^2x)
Simplify the expression in the denominator by taking the reciprocal of the fraction:
LHS: 1 + cos^2x / sin^2x
Combine the terms by finding a common denominator:
LHS: sin^2x / sin^2x + cos^2x / sin^2x
Simplify further:
LHS: (sin^2x + cos^2x) / sin^2x
Recall that sin^2x + cos^2x = 1, so replace this in the numerator:
LHS: 1 / sin^2x
Which is equivalent to the right side of the equation:
RHS: 1 / sin^2x
Therefore, the identity 1 + 1/tan^2x = 1/sin^2x is proven.
2) Start with the left side:
LHS: 2sin^2x - 1
Apply the identity sin^2x + cos^2x = 1 to the first term:
LHS: (1 - cos^2x) - 1
Distribute the negative sign:
LHS: 1 - cos^2x - 1
Simplify:
LHS: -cos^2x
Since cos^2x = 1 - sin^2x, replace cos^2x with this expression:
LHS: -(1 - sin^2x)
Simplify further:
LHS: -1 + sin^2x
Which is equivalent to the right side:
RHS: sin^2x - cos^2x
Therefore, the identity 2sin^2x - 1 = sin^2x - cos^2x is proven.
3) Start with the left side:
LHS: 1/cosx - cosx
Multiply the first term by cosx/cosx to get a common denominator:
LHS: cosx/cosx - cos^2x / cosx
Combine the terms:
LHS: (cosx - c