If sin A = 1/2 and cos B = √3/2, Find tan(A+B) and tan(A−B)

Question

If \[ \sin A=\frac{1}{2} \] and \[ \cos B=\frac{\sqrt{3}}{2} \] where \[ \frac{\pi}{2} < A < \pi \] and \[ 0 < B < \frac{\pi}{2} \] find the following:

(i) \(\tan(A+B)\)
(ii) \(\tan(A-B)\)

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, \[ \cos B=\frac{\sqrt{3}}{2} \]

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

\[ \sin B=\sqrt{1-\left(\frac{\sqrt{3}}{2}\right)^2} \]

\[ =\sqrt{1-\frac{3}{4}} \]

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

\[ \sin B=\frac{1}{2} \]

Since \[ 0 < B < \frac{\pi}{2} \] B lies in the first quadrant, so sine is positive.

Now,

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

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

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

(i) Find \(\tan(A+B)\)

Using formula:

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

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

\[ =\frac{0}{1+\frac{1}{3}} \]

\[ =0 \]

Therefore,

\[ \boxed{\tan(A+B)=0} \]

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

Using formula:

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

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

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

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

\[ =-\frac{2}{\sqrt{3}}\times\frac{3}{2} \]

\[ =-\sqrt{3} \]

Therefore,

\[ \boxed{\tan(A-B)=-\sqrt{3}} \]