Prove that the following is not an identity: `sec^2x + csc^2x = 1`

Identity is a statement that is true for any value of x. So, to prove that the given statement is not an identity, we can show that it is not true for at least one value of x. In fact, the given statement cannot possibly be true for any value of x.

First, let's rewrite the left side using reciprocal identities:

`secx = 1/cosx` and `cscx = 1/sinx` .

The left side becomes

`1/(cos^2x) +...

Identity is a statement that is true for any value of x. So, to prove that the given statement is not an identity, we can show that it is not true for at least one value of x. In fact, the given statement cannot possibly be true for any value of x.

First, let's rewrite the left side using reciprocal identities:

`secx = 1/cosx` and `cscx = 1/sinx` .

The left side becomes

`1/(cos^2x) + 1/(sin^2x)` .

Finding the common denominator and adding two fractions results in

`(sin^2x + cos^2x)/(cos^2x*sin^2x) = 1/(cos^2x*sin^2x)`  (Pythagorean identity, `cos^2x + sin^2x = 1`

is applied here.)

Finally, the denominator can be written as a single trigonometric function using the double-angle identity `sin(2x) = 2sinxcosx` :

`1/(cos^2x*sin^2x) = 1/((cosx*sinx)^2) =1/((sin(2x)/2)^2) = 4/(sin^2(2x))`

According to the given statement, this would have to be equal to 1:

`4/(sin^2(2x)) = 1` , which would mean that `sin^2(2x) = 4` , or that

`sin(2x) = +-2`

However, this cannot possibly be true because sine function, by definition, has to be always no more than 1 and no less than negative 1: `-1<=sinx<=1` .

There is no value of x for which sin(2x) = 2 or sin(2x) = -2.

So, not only `sec^2x + csc^2x = 1` is not an identity, it can never be true.