Answer
Required trigonometric functions are-
$\sin A = \frac{\sqrt 3}{2}$
$\\csc A = \frac{2}{\sqrt 3}$
$\\cos A = \frac{1}{2}$
$\\sec A = 2 $
$\\tan A = \sqrt 3 $
$\\\cot A = \frac{1}{\sqrt 3}$
Work Step by Step
Steps to Answer-
From the given diagram of triangle ABC, we will use given information and Pythagoras Theorem to solve for 'c'-
We know that -
$c^{2} =a^{2} + b^{2}$ ( Pythagoras Theorem)
Therefore -
$a^{2} =c^{2} - b^{2}$
$a^{2} = (2x)^{2} - x^{2}$
$a^{2} =4x^{2} - x^{2}$
$a^{2} = 3 x^{2}$
therefore $ a = x \sqrt 3$
Now we can write the required six T-functions of A using $a= x \sqrt 3$ ,$ b = x $ and c = $2x$
$\sin A = \frac{a}{c} = \frac{x\sqrt 3}{2x} = \frac{\sqrt 3}{2}$
$\\csc A = \frac{c}{a} =\frac{2x}{x\sqrt 3} = \frac{2}{\sqrt 3}$
$\\cos A = \frac{b}{c} =\frac{x}{2x} = \frac{1}{2}$
$\\sec A = \frac{c}{b} = \frac{2x}{x} = \frac{2}{1} = 2 $
$\\tan A = \frac{a}{b} = \frac{x\sqrt 3}{x} = \frac{\sqrt 3}{1} = \sqrt 3 $
$\\\cot A = \frac{b}{a} = \frac{x}{x\sqrt 3} = \frac{1}{\sqrt 3}$
