If $\tan\{\cos^{-1}(\frac{4}{5})+\tan^{-1}(\frac{2}{3})\}=\frac{a}{b}$, where a and b are co-prime natural numbers, then: |
a + b = 23 a – b = 12 3b = a + 2 2a = 3b |
a + b = 23 |
$cos^{-1}(\frac{4}{5})+\tan^{-1}(\frac{2}{3})= tan^{-1}(\frac{3}{4})+\tan^{-1}(\frac{2}{3})$ ⇒$\frac{a}{b}=tan[tan^{-1}(\frac{3}{4})+tan^{-1}(\frac{2}{3})]$ $=\tan[\tan^{-1}(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}×\frac{2}{3}})]=\tan[\tan^{-1}(\frac{17}{6})]=\frac{17}{6}$ ⇒ a = 17, b = 6 |