Solution :
The value of sec 45 degrees is \(\sqrt{2}\).
Proof :
Let ABC be a triangle, right angled at B, in which \(\angle\) A = \(\angle\) C = 45 degrees
\(\therefore\) BC = AB
Let AB = BC = a
Then by pythagoras theorem,
\(AC^2\) = \(AB^2\) + \(BC^2\) = \(a^2\) + \(a^2\) = \(2a^2\)
\(\implies\) AC = \(\sqrt{2}a\)
In \(\Delta\) ABC, \(\angle\) C = 45 degrees
By using trigonometric formulas,
\(sec 45^{\circ}\) = \(hypotenuse\over base\) = \(h\over b\)
\(sec 45^{\circ}\) = hypotenuse/side adjacent to 45 degrees = \(AC\over BC\) = \(\sqrt{2}a\over a\) = \(\sqrt{2}\)
Hence, the value of \(sec 45^{\circ}\) = \(\sqrt{2}\)
