Answer
$\sin A = \frac{\sqrt 7}{4}$ , $\\csc A = \frac{4}{\sqrt 7}$
$\\cos A = \frac{3}{4}$ , $\\sec A = \frac{4}{3}$
$\\tan A = \frac{\sqrt 7}{3}$ , $\\\cot A = \frac{3}{\sqrt 7}$
Work Step by Step
Steps to Answer-
First we will draw the diagram of triangle ABC and then use the given information and Pythagoras Theorem to solve for 'a'-
We know that -
$c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem)
Therefore -
$a^{2} =c^{2} - b^{2}$
$a^{2} = 4^{2} - 3^{2}$
$a^{2} = 16 - 9$
$a^{2} = 7$
therefore $ a = \sqrt 7$
Now we can write the required six T-functions of A using $a= \sqrt 7$ , b = 3 and c = 4
$\sin A = \frac{a}{c} = \frac{\sqrt 7}{4}$
$\\csc A = \frac{c}{a} =\frac{4}{\sqrt 7}$
$\\cos A = \frac{b}{c} = \frac{3}{4}$
$\\sec A = \frac{c}{b} = \frac{4}{3}$
$\\tan A = \frac{a}{b} =\frac{\sqrt 7}{3}$
$\\\cot A = \frac{b}{a} = \frac{3}{\sqrt 7}$
