Question

find sin(α) and cos(β), tan(α) and cot(β), and sec(α) and csc(β). (a) sin(α) and cos(β) (b) tan(α) and cot(β) (c) sec(α) and csc(β)

Explanation:

Step1: Find the adjacent side to $\alpha$ using Pythagorean theorem

Let the adjacent side to $\alpha$ be $x$. Then $x=\sqrt{21^{2}-12^{2}}=\sqrt{441 - 144}=\sqrt{297}=3\sqrt{33}$.

Step2: Calculate $\sin(\alpha)$

By the definition of sine in a right - triangle $\sin(\alpha)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{12}{21}=\frac{4}{7}$.

Step3: Calculate $\cos(\beta)$

Since $\alpha+\beta = 90^{\circ}$, $\cos(\beta)=\sin(\alpha)=\frac{4}{7}$.

Step4: Calculate $\tan(\alpha)$

$\tan(\alpha)=\frac{\text{opposite}}{\text{adjacent}}=\frac{12}{3\sqrt{33}}=\frac{4}{\sqrt{33}}=\frac{4\sqrt{33}}{33}$.

Step5: Calculate $\cot(\beta)$

Since $\alpha+\beta = 90^{\circ}$, $\cot(\beta)=\tan(\alpha)=\frac{4\sqrt{33}}{33}$.

Step6: Calculate $\sec(\alpha)$

$\sec(\alpha)=\frac{1}{\cos(\alpha)}$, and $\cos(\alpha)=\frac{3\sqrt{33}}{21}=\frac{\sqrt{33}}{7}$, so $\sec(\alpha)=\frac{7}{\sqrt{33}}=\frac{7\sqrt{33}}{33}$.

Step7: Calculate $\csc(\beta)$

Since $\alpha+\beta = 90^{\circ}$, $\csc(\beta)=\sec(\alpha)=\frac{7\sqrt{33}}{33}$.

Answer:

(a) $\sin(\alpha)=\frac{4}{7}$, $\cos(\beta)=\frac{4}{7}$
(b) $\tan(\alpha)=\frac{4\sqrt{33}}{33}$, $\cot(\beta)=\frac{4\sqrt{33}}{33}$
(c) $\sec(\alpha)=\frac{7\sqrt{33}}{33}$, $\csc(\beta)=\frac{7\sqrt{33}}{33}$