Question

1.      Given that csc A = , A in QI, and sec B = , sin B < 0, find

a.      cos (A – B)

b.     tan (A – B)

Answer 1

$$\csc A=\dfrac{1}{\sin A}=a, a\geq 1$$

$$\sin A=\dfrac{1}{a},$$

$$\cos A=\sqrt{1-\sin^2 A}=\sqrt{1-(\dfrac{1}{a})^2}=\dfrac{\sqrt{a^2-1}}{a}$$

$$\sec B=\dfrac{1}{\cos B}=b, |b|\geq 1$$

$$\cos B=\dfrac{1}{b}$$

$$\sin B=-\sqrt{1-\cos^2 B}=-\sqrt{1-(\dfrac{1}{b})^2}=-\dfrac{\sqrt{b^2-1}}{|b|}$$

a.

$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$

$$=\dfrac{\sqrt{a^2-1}}{a}(\dfrac{1}{b})+\dfrac{1}{a}(-\dfrac{\sqrt{b^2-1}}{|b|})$$

$$b\geq1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}-\sqrt{b^2-1}}{ab}$$

$$b\leq-1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}+\sqrt{b^2-1}}{ab}$$

b.

$$\tan(A-B)=\dfrac{\sin(A-B)}{\cos(a-B)}$$

$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$

$$=\dfrac{1}{a}(\dfrac{1}{b})-\dfrac{\sqrt{a^2-1}}{a}(-\dfrac{\sqrt{b^2-1}}{|b|})$$

$$b\geq1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}-\sqrt{b^2-1}}{ab}$$

$$\sin(A-B)=\dfrac{1+\sqrt{(a^2-1)(b^2-1)}}{ab}$$

$$\tan(A-B)=\dfrac{1+\sqrt{(a^2-1)(b^2-1)}}{\sqrt{a^2-1}-\sqrt{b^2-1}}$$

$$b\leq-1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}+\sqrt{b^2-1}}{ab}$$

$$\sin(A-B)=\dfrac{1-\sqrt{(a^2-1)(b^2-1)}}{ab}$$

$$\tan(A-B)=\dfrac{1-\sqrt{(a^2-1)(b^2-1)}}{\sqrt{a^2-1}+\sqrt{b^2-1}}$$