If =1 2 , =12 13 , where 2 <A< and 3 2 <B<2 find (A-B).

Question

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

✓

Answer

We have,
$\sin\text{A}=\frac{1}{2},$ and $\cos\text{B}=\frac{12}{13}$
$\therefore\cos\text{A}=\sqrt{1-\sin^2\text{A}}$ and $\sin\text{B}=-\sqrt{1-\cos^2\text{B}}$
[Cosin is negative in second quadrant and sine is negative in fourth quadrant]
$\Rightarrow\cos\text{A}=-\sqrt{1-\Big(\frac{1}{2}\Big)^2}$ and $\sin\text{B}=-\sqrt{1-\Big(\frac{12}{13}\Big)^2}$
$\Rightarrow\cos\text{A}=-\sqrt{1-\frac{1}{4}}$ and $\sin\text{B}=-\sqrt{1-\frac{144}{169}}$
$\Rightarrow\cos\text{A}=-\sqrt\frac{3}{4}$ and $\sin\text{B}=-\sqrt\frac{25}{169}$
$\Rightarrow\cos\text{A}=-\sqrt\frac{3}{2}$ and $\sin\text{B}-\frac{5}{13}$
$\therefore\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}=\frac{\frac{1}{2}}{\frac{-\sqrt{3}}{2}}=\frac{-1}{\sqrt{3}}$
$\tan\text{B}=\frac{\sin\text{B}}{\cos\text{B}}=\frac{-\frac{5}{13}}{\frac{{12}}{13}}=\frac{-5}{12}$
Now, $\tan\text{(A+B)}=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}\tan\text{B}}$
$=\frac{\frac{-1}{\sqrt{3}}-\Big(\frac{-5}{12}\Big)}{1+\Big(\frac{-1}{\sqrt{3}}\Big)\times\Big(\frac{-5}{12}\Big)}$
$=\frac{\frac{-1}{\sqrt{3}}+\frac{5}{12}}{1+\frac{5}{12\sqrt{3}}}$
$=\frac{\frac{-12+5\sqrt{3}}{12\sqrt{3}}}{\frac{12\sqrt{3}+5}{12\sqrt{3}}}$
$=\frac{5\sqrt{3}-12}{5+12\sqrt{3}}$
$\Rightarrow\tan\text{(A+B)}=\frac{5\sqrt{3}-12}{5+12\sqrt{3}}$