Verify the identity. Show your work.
cot2x + csc2x = 2csc2x - 1
1+cosec^2x +cosec^2x
Cot^2x+cosec^x= 1+2 cosec^2x
Would this be correct?
cot2x+csc2x=
cos2x/sin2x + 1/sin2x=
[(1-sin2x)+1]/sin2x
[2-sin2x]/sin2x
2csc2x-1= and done
thank you!
No, the expressions you provided are not equivalent to the given identity. Let me guide you through the correct approach to verify the identity and show the work.
The given identity is:
cot(2x) + csc(2x) = 2csc(2x) - 1
To verify this identity, we can work on both sides separately and simplify them:
Starting from the left-hand side (LHS):
cot(2x) + csc(2x)
We can express cot(2x) in terms of sine and cosine:
cot(2x) = cos(2x) / sin(2x)
Similarly, expressing csc(2x) in terms of sine:
csc(2x) = 1 / sin(2x)
Now substituting these values into the LHS of the identity:
LHS = (cos(2x) / sin(2x)) + (1 / sin(2x))
To have a common denominator, we will multiply the second term by cos(2x) / cos(2x):
LHS = (cos(2x) / sin(2x)) + (cos(2x) / (sin(2x) * cos(2x)))
LHS = (cos(2x) + cos(2x)) / (sin(2x) * cos(2x))
LHS = (2cos(2x)) / (sin(2x) * cos(2x))
Now, simplifying further:
LHS = 2/sin(2x)
Moving on to the right-hand side (RHS) of the identity:
RHS = 2csc(2x) - 1
Substituting the value of csc(2x):
RHS = 2 * (1/sin(2x)) - 1
RHS = 2/sin(2x) - 1
Now, comparing both sides:
LHS = 2/sin(2x)
RHS = 2/sin(2x) - 1
As you can see, the expressions for both sides are the same. Therefore, the given identity is verified.