Answer
$\sin A = \frac{3}{\sqrt 13}$
$\cos A = \frac{2}{\sqrt 13}$
$\tan A =\frac{3}{2}$
$\csc A = \frac{\sqrt 13}{3}$
$\sec A = \frac{\sqrt 13}{2}$
$\cot A = \frac{2}{3}$
Work Step by Step
Steps to Answer-
We will use given data about triangle ABC & Pythagoras Theorem to solve for 'c'-
We know that -
$c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem)
$c^{2} = (3)^{2} + (2)^{2}$
$c^{2} = 9 +4$
$c^{2} =13$
therefore $ c = \sqrt 13$
Now we can write the required T-functions of A using $a=3$ , b =2 and $ c = \sqrt 13$
$\sin A = \frac{a}{c} = \frac{3}{\sqrt 13}$
$\cos A = \frac{b}{c} = \frac{2}{\sqrt 13}$
$\tan A = \frac{a}{b} =\frac{3}{2}$
$\csc A = \frac{c}{a} = \frac{\sqrt 13}{3}$
$\sec A = \frac{c}{b} = \frac{\sqrt 13}{2}$
$\cot A = \frac{b}{a} = \frac{2}{3}$
