Which of the following is not an identity? a: (cosx+sinx)^2=1+2cosx sinx b: 1+cos^2x/sin^2=1 c: tan^2x=sec^2x-sin^2x-cos^2x d: (tanx+cotx)^2=csc^2x+sec^2x

Part A
(cosx+sinx)² = 1 + 2cosxsinx
Foil the left side.
cos²x + 2cosxsinx + sin²x = 1 + 2cosxsinx
Subtract 2cosxsinx from each side.
cos²x + sin²x = 1
And we're left with the Pythagorean identity!
Looks like it's not A.

Part B
( 1 + cos²x ) / sin²x = 1
Multiply each side by sin²x.
1 + cos²x = sin²x
Immediately, you should see a problem with this, and here's why:
Our Pythagorean identity from earlier can be put like this:
sin²x = 1 - cos²x
Substitute 1 - cos²x for sin²x in our earlier equation, and you get
1 + cos²x = 1 - cos²x
Which is clearly incorrect!
(If you really needed to verify this...subtract 1, divide by cos²x to get 1 = -1)
B is not an identity.

Part C
tan²x = sec²x - sin²x - cos²x
You might be able to recognize something from our Pythagorean identity.
Factor out a -1 from -sin²x - cos²x...
tan²x = sec²x -1(sin²x + cos²x)
Using our Pythagorean identity, 1 = sin²x + cos²x...
tan²x = sec²x - 1
Hey, look! Another trig identity! Don't recognize it? here...
sin²x/cos²x = 1/cos²x - 1
Multiply by cos²x...
sin²x = 1 - cos²x
There's also a great diagram for remembering the basic trig identities (besides double angle, half angle, angle addition and subtraction) called the magic hexagon...look it up, it's a great tool!

Part D
(tanx+cotx)² = csc²x + sec²x
This one sucks to prove but we can always use our last resort: expressing in terms of sine and cosine.
(sinx/cosx + cosx/sinx)² = (1/sinx)² + (1/cosx)²
Get the fractions on the left into a common denominator sinxcosx.
(sin²x/sinxcosx + cos²x/sinxcosx)² = (1/sinx)² + (1/cosx)²
(sin²x+cos²x/sinxcosx)² = (1/sinx)² + (1/cosx)²
(1/sinxcosx)² = (1/sinx)² + (1/cosx)²
Get the fractions on the right into a common denominator.
(1/sinxcosx)² = 1/sin²x + 1/cos²x
(1/sinxcosx)² = cos²x/sin²xcos²x + sin²x/sin²xcos²x
(1/sinxcosx)² = (sin²x+cos²x)/sin²xcos²x
(1/sinxcosx)² = 1/sin²xcos²x
(1/sinxcosx)² = (1/sinxcosx)²

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