Solution:

Given, A/cosA=b/cosB=C/cosC

Let b=2

Then, A/cosA=2/cosB=C/cosC

Multiplying all the terms, we get:

A×C×2=cosA×cosB×cosC

Area of the triangle = 1/2×b×h

We know, b=2

Let h be the perpendicular drawn from A to BC.

h=c×sinA

Now, we need to find the value of c.

Using the cosine rule in triangle ABC, we get:

c²=a²+b²-2ab×cosC

Substituting the values, we get:

c²=A²+4-4AcosC/cosA

Now, using the formula of area of a triangle, we get:

Area = 1/2×2×c×sinA

Substituting the values of c and sinA, we get:

Area = 2sinA√(A²+4-4AcosC/cosA)

Now, we have to find the value of sinA.

Using the sine rule in triangle ABC, we get:

a/sinA=b/sinB=c/sinC

Substituting the values, we get:

A/sinA=2/sinB=C/sinC

Now, multiplying all the terms, we get:

A×C×2=sinA×sinB×sinC

Simplifying, we get:

sinA=sqrt[(A²+4-4AcosC/cosA)/(A²+4+4AcosC/cosA)]

Substituting the value of sinA, we get:

Area = 2√[(A²+4-4AcosC/cosA)×(1-cos²C)/(cosA×cosC)]