Verify the trigonometric identity:
(1+sin x)/ (1-sinx)^2=(1+sin x)/[cos x]
Hint: [cos x] means the absolute value of cos x
Not sure what the little  characters mean, but unless
(1-sinx)^2 = |cosx|
it can't be true.
To verify the given trigonometric identity, we need to simplify both sides of the equation and show that they are equal.
Let's start with the left side of the equation:
(1 + sin x)/(1 - sin x)^2
To simplify it further, we can multiply the numerator and denominator by (1 - sin x):
[(1 + sin x)/(1 - sin x)] * [(1 - sin x)/(1 - sin x)]
= (1 + sin x)(1 - sin x) / (1 - sin x)^2
Using the difference of squares formula, (a^2 - b^2) = (a + b)(a - b), we can simplify the numerator:
= (1 - sin^2x) / (1 - sin x)^2
= cos^2x / (1 - sin x)^2
Now, let's simplify the right side of the equation:
(1 + sin x)/[cos x]
To make it easier, let's express it as (1 + sin x)*(1/cos x):
= (1 + sin x) / cos x
To compare it with the left side of the equation, we need to rationalize the denominator by multiplying the numerator and denominator by (1 - sin x):
[(1 + sin x)*(1 - sin x)] / [cos x * (1 - sin x)]
= (1 - sin^2x) / [cos x * (1 - sin x)]
= cos^2x / (cos x * (1 - sin x))
= cos x / (1 - sin x)
Now, you can see that the right side of the equation is equal to cos x / (1 - sin x).
Therefore, we have successfully verified the trigonometric identity:
(1 + sin x)/(1 - sin x)^2 = (1 + sin x)/[cos x]