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.