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

Question

Oliver Atkinson · Accepted Answer

This is true #tan p+cot p=2 csc 2p# see the explanation

the given #tan p+cot p=2 csc 2p# start from left side #sin p/cos p+cos p/sin p=2 csc 2p# #sin p/cosp*sinp/sinp+cos p/sin p *cos p/cos p=2 csc 2p# #sin^2 p/(sin p cos p)+cos^2p/(sin p cos p)=2 csc 2p# from #sin^2p+cos^2p=1# equation becomes #(sin^2 p+cos^2p)/(sin p cos p)=2 csc 2p# #1/(sin p cos p)=2 csc 2p# #1/(sin p cos p)*2/2=2 csc 2p# #2/(2sin p cos p)=2 csc 2p# From #sin 2p=2 sin p cos p# equation becomes #2/(sin 2p)=2 csc 2p# #2*(1/(sin 2p))=2 csc 2p# Take note: #csc 2p=1/(sin 2p)# #2 csc 2p=2 csc 2p#

James Adkins · Answer

undefined 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).