5. If sec α = 2/√3, , then find the value of (1-cosec α)/(1+cosec α) , where α is in IV quadrant.

Solution: secα = 2/√3   [Given]

But we know that,   secα = r/x  = 2/√3

∴  r = 2 and x = √3

Now, we know that,

r²  = x²   + y²

∴  2² = (√3)² + y²

∴  4 = 3 + y²

∴ 4 – 3 = y²

∴  y² = 1

∴  y = ±√1

∴  y = ± 1

Since, α is in IV quadrant,

∴  x is positive but y is negative,

∴  y = - 1

Now,

cosec α = r/y   = -2/1

∴ cosec α = -2

Now,

(1 – cosecα)/(1 + cosec α)  = [1 – (-2) ]/[1+(-2) ]

∴ (1- cosec α)/(1+cosec α)=(1+2)/(1-2)

∴ (1- cosec α)/(1+cosec α)= 3/(-1)

∴ (1-cosec α)/(1+cosec α)= - 3