Answer :

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

sec =

=

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

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 *k*^{2} = 144 *k*^{2} + BC^{2}

25 *k*^{2} = BC^{2}

BC = 5 *k*

Sin =

= =

Cos θ =

= =

Tan θ =

= =

Cot θ =

= =

Cosec θ =

= =

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