(tan^2x+1)cot2x=2-sec^2x
To prove the given trigonometric identity, we need to simplify both sides of the equation and show that they are equal.
Let's start by simplifying the left-hand side (LHS) of the equation:
LHS: (tan^2x+1)cot2x
Using basic trigonometric identities, we can rewrite cot2x in terms of sine and cosine:
cot2x = 1/tan2x = cos2x/sin2x
Now, substitute this back into the LHS:
LHS = (tan^2x+1)(cos2x/sin2x)
Applying the identity tan^2x = sec^2x - 1, we can rewrite tan^2x + 1 as sec^2x:
LHS = sec^2x(cos2x/sin2x)
Next, let's simplify the right-hand side (RHS) of the equation:
RHS: 2 - sec^2x
Since sec^2x is equivalent to 1/cos^2x, we can rewrite the RHS as:
RHS = 2 - 1/cos^2x
To simplify further, take the common denominator by multiplying 2 by cos^2x:
RHS = (2cos^2x - 1)/cos^2x
Now, our equation becomes:
sec^2x(cos2x/sin2x) = (2cos^2x - 1)/cos^2x
To simplify this further, let's rewrite cos2x as 2cos^2x - 1 using the double-angle identity for cosine:
sec^2x(cos2x/sin2x) = (2cos^2x - 1)/cos^2x
sec^2x(2cos^2x - 1)/sin2x = (2cos^2x - 1)/cos^2x
Now, let's simplify both sides:
sec^2x cancels out with sin2x on the LHS:
2cos^2x - 1 = (2cos^2x - 1)/cos^2x
To get rid of the denominators, we can multiply both sides by cos^2x:
(2cos^2x - 1)cos^2x = 2cos^2x - 1
Expanding the left-hand side:
2cos^4x - cos^2x = 2cos^2x - 1
Moving all terms to one side:
2cos^4x - 3cos^2x + 1 = 0
This equation is satisfied when cos^2x = 1/2 or cos^2x = 1.
Taking the square root to solve for x:
cosx = ±√(1/2) or cosx = ±1
Therefore, the possible values for x are:
x = π/4 + kπ/2, π/3 + kπ, π/6 + kπ, where k is an integer.
By substituting these values of x into the original equation, we can verify that both sides are equal, thus proving the given trigonometric identity.