1. Given that csc A = , A in QI, and sec B = , sin B < 0, find
a. cos (A – B)
b. tan (A – B)
"\\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.
"=\\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.
"\\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}"
"\\tan(A-B)=\\dfrac{1+\\sqrt{(a^2-1)(b^2-1)}}{\\sqrt{a^2-1}-\\sqrt{b^2-1}}"
"\\tan(A-B)=\\dfrac{1-\\sqrt{(a^2-1)(b^2-1)}}{\\sqrt{a^2-1}+\\sqrt{b^2-1}}"
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