Question

If $\cos A=\frac{1}{2}$ and $\sin B=\frac{1}{\sqrt{2}}$, find the value of : $\frac{\tan A-\tan B}{1+\tan A\tan B}$

Answer 1

$cosA=\frac{1}{2},sinB=\frac{1}{\sqrt{2}}$

$Si{n}^{2}A+co{s}^{2}A=1$ $si{n}^{2}B+co{s}^{2}B=1$

$si{n}^{2}A+\frac{1}{4}=1$ $\frac{1}{2}+co{s}^{2}B=1$

$si{n}^{2}A=1-1/4$ $co{s}^{2}B=1/2$

$si{n}^{2}A=3/4$ $cosB=1/\sqrt{2}$

$sinA=\frac{\sqrt{3}}{2}$ $tanB=\frac{sinB}{cosB}$

$tanA=\frac{sinA}{cosA}$ $=\frac{1/\sqrt{2}}{1/\sqrt{2}}$

$tanA=\frac{\sqrt{3}/2}{1/2}$ tanB = 1

$tanA=\sqrt{3}$

$\therefore \frac{tanA-tanB}{1+tanAtanB}$ $\frac{\sqrt{3}-1}{1+\sqrt{3}}$ $=\frac{\sqrt{3}-1}{\sqrt{3}+1}$