If sin A = 1/2 and cos B = 12/13, Find tan(A−B)

Question

If \[ \sin A=\frac{1}{2} \] and \[ \cos B=\frac{12}{13} \] where \[ \frac{\pi}{2} < A < \pi \] and \[ \frac{3\pi}{2} < B < 2\pi \] find:

\[ \tan(A-B) \]

Solution

Given: \[ \sin A=\frac{1}{2} \]

Using \[ \sin^2 A+\cos^2 A=1 \]

\[ \cos A=\sqrt{1-\left(\frac{1}{2}\right)^2} \]

\[ =\sqrt{1-\frac{1}{4}} \]

\[ =\sqrt{\frac{3}{4}} \]

\[ \cos A=\frac{\sqrt{3}}{2} \]

Since \[ \frac{\pi}{2} < A < \pi \] A lies in the second quadrant, where cosine is negative.

Therefore, \[ \cos A=-\frac{\sqrt{3}}{2} \]

Now,

\[ \tan A=\frac{\sin A}{\cos A} \]

\[ =\frac{1/2}{-\sqrt{3}/2} \]

\[ =-\frac{1}{\sqrt{3}} \]

Also, \[ \cos B=\frac{12}{13} \]

Using \[ \sin^2 B+\cos^2 B=1 \]

\[ \sin B=\sqrt{1-\left(\frac{12}{13}\right)^2} \]

\[ =\sqrt{1-\frac{144}{169}} \]

\[ =\sqrt{\frac{25}{169}} \]

\[ \sin B=\frac{5}{13} \]

Since \[ \frac{3\pi}{2} < B < 2\pi \] B lies in the fourth quadrant, where sine is negative.

Therefore, \[ \sin B=-\frac{5}{13} \]

Now,

\[ \tan B=\frac{\sin B}{\cos B} \]

\[ =\frac{-5/13}{12/13} \]

\[ =-\frac{5}{12} \]

Find \(\tan(A-B)\)

Using formula:

\[ \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \]

\[ =\frac{-\frac{1}{\sqrt{3}}-\left(-\frac{5}{12}\right)}{1+\left(-\frac{1}{\sqrt{3}}\times-\frac{5}{12}\right)} \]

\[ =\frac{-\frac{1}{\sqrt{3}}+\frac{5}{12}}{1+\frac{5}{12\sqrt{3}}} \]

Taking LCM in numerator:

\[ =\frac{\frac{-12+5\sqrt{3}}{12\sqrt{3}}}{\frac{12\sqrt{3}+5}{12\sqrt{3}}} \]

\[ =\frac{-12+5\sqrt{3}}{12\sqrt{3}+5} \]

Therefore,

\[ \boxed{\tan(A-B)=\frac{-12+5\sqrt{3}}{12\sqrt{3}+5}} \]