Answer 1

We know `sin theta = "𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒"/"ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒"`

Let x be the adjacent side

By applying Pythagoras theorem

𝐴𝐶² = 𝐴𝐵² + 𝐵𝐶²

b² = a² + x²

x² = b² − a²

`x = sqrt(b^2 - a^2)`

`sec theta = (AB)/(BC) = b/(sqrt(b^2 - a^2))`

`tan theta = (AB)/(BC) = a/(sqrt(b^2 - a^2))`

`sec theta + tan theta = b/(b^2 - a^2) + a/(sqrt(b^2 - a^2))`

`= (b + a)/(sqrt(b^2 - a^2)) = (b+ a)/sqrt((b + a)(b - a)) = (b + a)/sqrt(b + a) - 1/(sqrt(b - a)) = sqrt((b + a)/(b - a))`