Given sec <

Consider a right-angle triangle ΔABC, right-angled at point B

sec =

=

If AC is 13k, AB will be 12k, where k is a positive integer.

Now according to Pythagoras theorem,

The square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Applying Pythagoras theorem in ΔABC, we obtain

(AC)² = (AB)² + (BC)²

(13 k)² = (12 k)² + (BC)²

169 k² = 144 k² + BC²

25 k² = BC²

BC = 5 k

Sin =

= =

Cos θ =

= =

Tan θ =

= =

Cot θ =

= =

Cosec θ =

= =