verify the identity:
1- (cos^2x)/(1-sinx)= -sinx
I already did this today, scroll down
verify the identity:
1-cos^2x/1-sinx= -sinx
Trigonometry - Damon, Monday, May 30, 2011 at 6:43pm
assume you mean (this took some serious detective work - please be careful with parentheses)
1- (cos^2x)/(1-sinx)= -sinx
cos^2x = 1 - sin^2 x
so we have
1 - (1-sin x)(1+sin x)/(1-sin x) = -sin x
1 - (1+sin x) = -sin x
1 - 1 - sin x = -sin x
-sin x = -sin x
To verify the identity, let's simplify the left-hand side (LHS) of the equation and see if it is equal to the right-hand side (RHS):
We start with the LHS:
LHS = (cos^2x)/(1 - sinx)
Using the identity cos^2x = 1 - sin^2x, we can rewrite the numerator:
LHS = (1 - sin^2x)/(1 - sinx)
Now, we can factor out (1 - sinx) from the numerator:
LHS = [(1 - sinx)(1 + sinx)]/(1 - sinx)
The (1 - sinx) terms cancel out:
LHS = 1 + sinx
Now, we can compare this expression to the RHS, which is -sinx:
RHS = -sinx
Since LHS = 1 + sinx and RHS = -sinx, we can see that the two sides are not equal.
Therefore, the given equation is not an identity and cannot be verified.
To verify the identity (cos^2x)/(1-sinx) = -sinx, we need to simplify both sides of the equation and show that they are equal.
Starting with the left side of the equation:
(cos^2x)/(1-sinx)
We can rewrite cos^2x as (1-sinx^2) using the Pythagorean identity: cos^2x = 1 - sin^2x.
(1 - sin^2x)/(1 - sinx)
Now we can factor out a negative sign from the numerator:
-(sin^2x - 1)/(1 - sinx)
Using the difference of squares identity: a^2 - b^2 = (a + b)(a - b), we can rewrite the numerator:
-((sinx + 1)(sinx - 1))/(1 - sinx)
Now we can cancel out the common factor of (1 - sinx) in the numerator and denominator:
-(sinx + 1)/(sinx - 1)
Now let's simplify the right side of the equation:
-sinx
Since the left side of the equation is already in simplified form, we can compare it with the right side:
-(sinx + 1)/(sinx - 1) = -sinx
So, we have successfully verified the given identity.