Question

If $sinA=\frac{1}{2},$ $cosB=\frac{12}{13},$ where $\frac{\pi }{2}<A<\pi$ and $\frac{3\pi }{2}<B<2\pi ,$ find tan(A-B)

Answer 1

We have,
$sinA=\frac{1}{2},cosB=\frac{12}{13}$
$\therefore cosA=\sqrt{1-si{n}^{2}A}$ and $sinB=-\sqrt{1-co{s}^{2}B}$
[ $\because$ cosine is negetive in second quadrant and sine is negative in fourth quadrant]
$\Rightarrow cosA=-\sqrt{1-{\left(\frac{1}{2}\right)}^2}$ and $sinB=-\sqrt{1-{\left(\frac{12}{13}\right)}^2}$
$\Rightarrow cosA=-\sqrt{1-\frac{1}{4}}$ and
$sinB=-\sqrt{1-\frac{144}{169}}$
$\Rightarrow cosA=-\sqrt{\frac{3}{4}}$ and
$sinB=-\sqrt{\frac{25}{169}}$
$\Rightarrow cosA=-\frac{\sqrt{3}}{2}$ and $sinB=\frac{5}{13}$
$\therefore tanA=\frac{sinA}{cosA}=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{-1}{\sqrt{3}}$
and, $tanB=\frac{sinB}{cosB}=\frac{-\frac{5}{13}}{\frac{12}{13}}=\frac{-5}{12}$
Now,
$tan(A-B)=\frac{tanA-tanB}{1+tanA.tanB}$
$=\frac{\frac{-1}{\sqrt{3}-\left(\frac{-5}{12}\right)}}{1+(-\frac{1}{\sqrt{3}})\times \left(\frac{-5}{12}\right)}=\frac{\frac{-1}{\sqrt{3}}-\frac{-5}{12}}{1+(-\frac{1}{\sqrt{3}})\times \left(\frac{-5}{12}\right)}$
$=\frac{\frac{-1}{\sqrt{3}}+\frac{5}{12}}{1+\frac{5}{12\sqrt{3}}}=\frac{\frac{-12+5\sqrt{3}}{12\sqrt{3}}}{\frac{12\sqrt{3}+5}{12\sqrt{3}}}$
= $\frac{5\sqrt{3}-12}{5\sqrt{3}+12\sqrt{3}}$
$\Rightarrow tan(A-B)=\frac{5\sqrt{3}-12}{5\sqrt{3}+12\sqrt{3}}$