Answer 1

Let x be, the hypotenuse

By Pythagoras we get

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

𝑥² = 144 + 169

`x = sqrt313`

`sin theta = (AB)/(AC) = 12/sqrt313`

`cos theta = (BC)/(AC) = 13/sqrt313`

Substitute, Sin 𝜃, cos 𝜃 in equation we get

`(2 sin theta cos theta)/(cos^2 theta - sin^2 theta) => (2 xx 12/sqrt313 xx 13/sqrt313)/(169/313 - 144/313)`

`= (312/313)/(25/313) = 312/25`