# How do you prove #tan p + cot p = 2 csc 2 p#?

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To prove tan(p) + cot(p) = 2csc(2p), we'll use the definitions of tangent, cotangent, and cosecant, along with trigonometric identities.

Given: tan(p) = sin(p)/cos(p) cot(p) = cos(p)/sin(p) csc(2p) = 1/sin(2p)

We know that sin(2p) = 2sin(p)cos(p) by the double-angle identity.

So, csc(2p) = 1/(2sin(p)cos(p))

Now, we'll rewrite tan(p) and cot(p) in terms of sin and cos:

tan(p) = sin(p)/cos(p) cot(p) = cos(p)/sin(p)

Adding them: tan(p) + cot(p) = (sin(p)/cos(p)) + (cos(p)/sin(p))

To combine these fractions, we'll get a common denominator: tan(p) + cot(p) = (sin^2(p) + cos^2(p))/(sin(p)cos(p))

Recall the Pythagorean identity: sin^2(p) + cos^2(p) = 1 So, the expression becomes: tan(p) + cot(p) = 1/(sin(p)cos(p))

Comparing with 2csc(2p), which we found to be 1/(2sin(p)cos(p)), we see they are equivalent.

Therefore, tan(p) + cot(p) = 2csc(2p).

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