Verify the Identity.
cos^4xcsc^2x = csc^2x + som^2x - 2
som is not a trig function, check your typing
I'm sorry I meant sin
To verify the identity, we will simplify the left-hand side (LHS) of the equation:
LHS: cos^4x * csc^2x
First, let's write cos^4x as (cos^2x)^2:
LHS: (cos^2x)^2 * csc^2x
Next, we can rewrite csc^2x as 1/sin^2x:
LHS: (cos^2x)^2 * (1/sin^2x)
Expanding the square of cos^2x:
LHS: (cos^2x * cos^2x) * (1/sin^2x)
Simplifying the product of cos^2x:
LHS: cos^4x * (1/sin^2x)
We can rewrite cos^4x as (1 - sin^2x)^2:
LHS: (1 - sin^2x)^2 * (1/sin^2x)
Next, let's expand the square:
LHS: (1 - 2sin^2x + sin^4x) * (1/sin^2x)
Multiplying both terms:
LHS: (1/sin^2x) - (2sin^2x/sin^2x) + (sin^4x/sin^2x)
Simplifying the second term:
LHS: (1/sin^2x) - 2 + (sin^4x/sin^2x)
We can rewrite 1/sin^2x as csc^2x:
LHS: csc^2x - 2 + (sin^4x/sin^2x)
Now, we can simplify the right-hand side (RHS) of the equation:
RHS: csc^2x + sin^2x - 2
Comparing the simplified LHS and RHS, we can see that they are the same:
LHS: csc^2x - 2 + (sin^4x/sin^2x)
RHS: csc^2x + sin^2x - 2
Therefore, we have verified that the given identity is true.