Question

If $\cos A=0.5$ and $\cos B=\frac{1}{\sqrt{2}}$; find the value of :- $\frac{\tan A-\tan B}{1+\tan A\tan B}$.

Answer 1

$\cos A=0.5\left(\text{Given}\right)$

$\Rightarrow \cos A=\frac{1}{2}$

$\Rightarrow \sec A=2$

Also,

$\cos B=\frac{1}{\sqrt{2}}\left(\text{Given}\right)$

$\Rightarrow \sec B=\sqrt{2}$

Now,

$\frac{\tan A-\tan B}{1+\tan A\tan B}$

$=\frac{\sqrt{{\sec }^{2}A-1}-\sqrt{{\sec }^{2}B-1}}{1+\sqrt{{\sec }^{2}A-1}\sqrt{{\sec }^{2}B-1}}$

$=\frac{\sqrt{3}-1}{1+\sqrt{3}}$

$=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}$

$=\frac{{(\sqrt{3}-1)}^2}{{\left(\sqrt{3}\right)}^2-1^2}$

$=\frac{3+1-2\sqrt{3}}{3-1}$

$=2-\sqrt{3}$