Answer
$\sin\theta=\dfrac{40}{41}$
$\cos\theta=\dfrac{9}{41}$
$\tan\theta=\dfrac{40}{9}$
$\csc\theta=\dfrac{41}{40}$
$\sec\theta=\dfrac{41}{9}$
$\cot\theta=\dfrac{9}{40}$
Work Step by Step
The triangle is shown in the attached image below.
Let $c$ be the unknown cathetus in the triangle. Use the Pythagorean Theorem to find $c$:
$c=\sqrt{41^{2}-40^{2}}=\sqrt{1681-1600}=\sqrt{81}=9$
$\sin\theta=\dfrac{opposite}{hypotenuse}=\dfrac{40}{41}$
$\cos\theta=\dfrac{adjacent}{hypotenuse}=\dfrac{9}{41}$
$\tan\theta=\dfrac{opposite}{adjacent}=\dfrac{40}{9}$
$\csc\theta=\dfrac{hypotenuse}{opposite}=\dfrac{41}{40}$
$\sec\theta=\dfrac{hypotenuse}{adjacent}=\dfrac{41}{9}$
$\cot\theta=\dfrac{adjacent}{opposite}=\dfrac{9}{40}$
