Answer
$a)$ $\sin\alpha=\cos\beta=\dfrac{4}{7}$
$b)$ $\tan\alpha=\cot\beta=\dfrac{4\sqrt{33}}{33}$
$c)$ $\sec\alpha=\csc\beta=\dfrac{7\sqrt{33}}{33}$
Work Step by Step
The triangle is shown below.
Let $c$ be the unknown cathetus of the triangle. Find it using the Pythagorean Theorem:
$c=\sqrt{7^{2}-4^{2}}=\sqrt{49-16}=\sqrt{33}$
$a)$ $\sin\alpha$ and $\cos\beta$
$\sin\alpha=\dfrac{opposite}{hypotenuse}=\dfrac{4}{7}$
$\cos\beta=\dfrac{adjacent}{hypotenuse}=\dfrac{4}{7}$
$b)$ $\tan\alpha$ and $\cot\beta$
$\tan\alpha=\dfrac{opposite}{adjacent}=\dfrac{4}{\sqrt{33}}=\dfrac{4\sqrt{33}}{33}$
$\cot\beta=\dfrac{adjacent}{opposite}=\dfrac{4}{\sqrt{33}}=\dfrac{4\sqrt{33}}{33}$
$c)$ $\sec\alpha$ and $\csc\beta$
$\sec\alpha=\dfrac{hypotenuse}{adjacent}=\dfrac{7}{\sqrt{33}}=\dfrac{7\sqrt{33}}{33}$
$\csc\beta=\dfrac{hypotenuse}{opposite}=\dfrac{7}{\sqrt{33}}=\dfrac{7\sqrt{33}}{33}$
