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URGENT!!

Show that sin ^-1 (4/5) +cos ^-1 (2/square root of 5)=cot ^-1 (2/11)

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$L e t \sin^{- 1} (\frac{4}{5}) = x \sin x = \frac{4}{5} \cos x = \sqrt{1 - {(\frac{4}{5})}^{2}} = \frac{3}{5} \tan x = \frac{4}{3} \Rightarrow x = \tan^{-1} (\frac{4}{3}) = \sin^{- 1} (\frac{4}{5}) ....... (i) N o w, \cos^{- 1} (\frac{2}{\sqrt{5}}) = y \cos y = \frac{2}{\sqrt{5}} \sin y = \sqrt{1 - {(\frac{2}{\sqrt{5}})}^{2}} = \frac{1}{\sqrt{5}} \tan y = \frac{1}{2} \Rightarrow x = \tan^{-1} (\frac{1}{2}) = \cos^{- 1} (\frac{2}{\sqrt{5}}) ....... (i i) L . H . S \sin^{- 1} (\frac{4}{5}) + \cos^{- 1} (\frac{2}{\sqrt{5}}) = \tan^{-1} (\frac{4}{3}) + \tan^{-1} (\frac{1}{2}) = \tan^{-1} \frac{\frac{4}{3} + \frac{1}{2}}{1 - \frac{4}{3} \times \frac{1}{2}} [F r o m (i) a n d (i i)] = \tan^{-1} \frac{\frac{11}{6}}{\frac{1}{3}} = \tan^{-1} \frac{11}{2} = \cot^{-1} \frac{2}{11} = R . H . S$

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