# need help in rewite the expression 8/(`x^(2)-4)` `^(3/2)` x=2 sec `theta`

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`8/(x^2-4)^(3/2)`

`x = 2sectheta`

`x^2 = 4sec^2theta`

`x^2-4 = 4sec^2theta-4 = 4(sec^2theta-1)`

From trigonometry we know that;

`1+tan^2theta = sec^2theta`

`sec^2theta-1 = tan^2theta`

`(x^2-4)^(3/2)`

`= (4(tan^2theta))^(3/2)`

`= 4^(3/2)xx(tan^2theta)^(3/2)`

`= (2^2)^(3/2)xx(tan^2theta)^(3/2)`

`= 8tan^3theta`

`8/(x^2-4)^(3/2)`

`= 8/(8tan^3theta)`

`= 1/(tan^3theta)`

** So the answer is** `8/(x^2-4)^(3/2) = 1/(tan^3theta)`

**Sources:**