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:
Answer & explanation
Correct answer: option 1
$cos^{-1}(\frac{4}{5})+\tan^{-1}(\frac{2}{3})= tan^{-1}(\frac{3}{4})+\tan^{-1}(\frac{2}{3})$
⇒$\frac{a}{b}=tan[tan^{-1}(\frac{3}{4})+tan^{-1}(\frac{2}{3})]$
$=\tan[\tan^{-1}(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}×\frac{2}{3}})]=\tan[\tan^{-1}(\frac{17}{6})]=\frac{17}{6}$
⇒ a = 17, b = 6