prove as an identity

1/1+sin(x) + 1/1-sin(x)

oops i forgot the =2sec^2

identities are equations.

I don't see an equation.

so it is 1/(1+sinx) + 1/(1-sinx) = 2sec^2 x

LS
= (1 - sinx + 1 + sinx)/((1+sinx)(1-sinx))
= 2/(1 - sin^2x)
= 2 cos^2x
= 2sec^2x
= RS

To prove the given expression as an identity, we need to simplify it to a single expression that is always true, regardless of the value of x.

Starting with the given expression:

1/(1 + sin(x)) + 1/(1 - sin(x))

To simplify this expression, we need to find a common denominator. The common denominator in this case is (1 + sin(x))(1 - sin(x)).

Now, let's rewrite each fraction with the common denominator:

[(1 - sin(x))/(1 + sin(x))(1 - sin(x))] + [(1 + sin(x))/(1 + sin(x))(1 - sin(x))]

Next, let's combine the numerators:

[(1 - sin(x)) + (1 + sin(x))]/[(1 + sin(x))(1 - sin(x))]

Simplifying the numerator:

1 - sin(x) + 1 + sin(x)]/[(1 + sin(x))(1 - sin(x))]

Now, combine like terms in the numerator:

2 / [(1 + sin(x))(1 - sin(x))]

Notice that (1 + sin(x))(1 - sin(x)) is equal to 1 - sin^2(x), which is the identity for the difference of squares:

(a + b)(a - b) = a^2 - b^2

Using this identity, we can simplify further:

2 / (1 - sin^2(x))

Since sin^2(x) = 1 - cos^2(x) (which is another identity), we can replace sin^2(x) with 1 - cos^2(x):

2 / (1 - (1 - cos^2(x)))

Simplifying the denominator:

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

2 / cos^2(x)

Finally, we can rewrite this expression using the reciprocal of cos^2(x), which is sec^2(x):

2 * sec^2(x)

Therefore, the original expression 1/(1 + sin(x)) + 1/(1 - sin(x)) simplifies to 2 * sec^2(x), and this shows that the given expression is an identity.