Given A = 3. Find the value of (i) A (ii) A (iii) A (iv) A (v) A (In Surd form)

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1. Given $\tan A = 3$ . Find the value of (i) $\sin A$ (ii) $\cos A$ (iii) $\sec A$ (iv) $\csc A$ (v) $\cot A$ (In Surd form)

Use the given $\tan A = 3$ to construct a right-angled triangle.
Since $\tan A = \frac{\text{opposite}}{\text{adjacent}}$ , we can consider the opposite side to be 3 units and the adjacent side to be 1 unit.
Use the Pythagorean theorem to find the hypotenuse.
Let the opposite side be $o=3$ and the adjacent side be $a=1$ . Let the hypotenuse be $h$ .
$h^2 = o^2 + a^2$ $h^2 = 3^2 + 1^2$ $h^2 = 9 + 1$ $h^2 = 10$ $h = \sqrt{10}$
Calculate each trigonometric ratio.
(i) $\sin A$
$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{\sqrt{10}}$
Rationalize the denominator:
$\sin A = \frac{3}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$
The final answer for (i) is $\boxed{\frac{3\sqrt{10}}{10}}$ .

(ii) $\cos A$
$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{10}}$
Rationalize the denominator:
$\cos A = \frac{1}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{10}}{10}$
The final answer for (ii) is $\boxed{\frac{\sqrt{10}}{10}}$ .

(iii) $\sec A$
$\sec A = \frac{1}{\cos A} = \frac{1}{\frac{1}{\sqrt{10}}} = \sqrt{10}$
The final answer for (iii) is $\boxed{\sqrt{10}}$ .

(iv) $\csc A$
$\csc A = \frac{1}{\sin A} = \frac{1}{\frac{3}{\sqrt{10}}} = \frac{\sqrt{10}}{3}$
The final answer for (iv) is $\boxed{\frac{\sqrt{10}}{3}}$ .

(v) $\cot A$
$\cot A = \frac{1}{\tan A} = \frac{1}{3}$
The final answer for (v) is $\boxed{\frac{1}{3}}$ .

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