Question

Prove the following,

$ta{n}^{-1}\frac{2}{11}+ta{n}^{-1}\frac{7}{24}=ta{n}^{-1}\frac{1}{2}$

Answer 1

Given $ta{n}^{-1}\frac{2}{11}+ta{n}^{-1}\frac{7}{24}=ta{n}^{-1}\frac{1}{2}$
LHS=$ta{n}^{-1}\left(\frac{2}{11}\right)+ta{n}^{-1}\left(\frac{7}{24}\right)=ta{n}^{-1}\left(\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}.\frac{7}{24}}\right)(\because ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right))=ta{n}^{-1}\left(\frac{\frac{48+77}{264}}{1-\frac{14}{264}}\right)=ta{n}^{-1}\left(\frac{\frac{125}{264}}{\frac{264-14}{264}}\right)=ta{n}^{-1}\frac{\frac{125}{264}}{\frac{250}{264}}=ta{n}^{-1}(\frac{125}{264}\times \frac{264}{250})=ta{n}^{-1}\frac{1}{2}=RHS$.

Hence proved.