Question

cos0 =root 3/2 then find value of 1-sec0 /1+ cosec0​

Answer 1

cosA =√3/2 = Base/Hypotenuse

by Pythagoras theorem

H^2 = P^2 + B^2

(2)^2 = P^2 + (√3)^2

4 = P^2 + 3

4-3 = P^2

P^2 = 1

P = 1

now, we have

Hypotenuse = 2

Base =√3

perpendicular = 1

sinA = perpendicular/Hypotenuse = 1/2

cosecA = 1/sinA = 2

cosA = Base/Hypotenuse = √3/2

secA = 1/cosA = 2/√3

Now,

(1-secA) / (1+cosecA)

(1 - 2/√3) / (1 + 2)

(√3 - 2)/√3 / 3

(√3 - 2)/3√3