Question

Find the value of $tan[cos^{-1}(\frac{4}{5}) + tan^{-1}(\frac{2}{3})]$.

Answer 1

hope it help you dear.......

Answer 2

Answer:

$tan[cos^{-1}(\frac{4}{5}) + tan^{-1}(\frac{2}{3})]=\frac{17}{6}$

Step-by-step explanation:

To find the value of

$tan[cos^{-1}(\frac{4}{5}) + tan^{-1}(\frac{2}{3})]$

if we covert cos inverse in terms of tan inverse than it will be easily solved

As we know that

$cos^{-1}x=tan^{-1}(\frac{\sqrt{1-x^{2}} }{x} )\\\\cos^{-1}(\frac{4}{5}) =tan^{-1}(\frac{\sqrt{1-(\frac{4}{5}) ^{2}} }{\frac{4}{5} } )\\\\\\=tan^{-1}(\frac{\sqrt{1-(\frac{16}{25})} }{\frac{4}{5} } )\\\\=tan^{-1}(\frac{3\times5}{5\times4})\\\\=tan^{-1}\frac{3}{4}$

since

$tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy} )\\\\\\tan^{-1}\frac{3}{4} +tan^{-1}\frac{2}{3} =tan^{-1}(\frac{\frac{3}{4} +\frac{2}{3} }{1-\frac{3}{4} \frac{2}{3}} )\\\\\\=tan^{-1}(\frac{17}{6} )\\\\so\\\\\\cos^{-1}(\frac{4}{5}) + tan^{-1}(\frac{2}{3})=tan^{-1}(\frac{17}{6} )\\\\\\tan[cos^{-1}(\frac{4}{5}) + tan^{-1}(\frac{2}{3})]=tan[tan^{-1}(\frac{17}{6} )]\\\\=(\frac{17}{6} )$