Let α=sin−1⁡3685,β=cos−1⁡45 and γ=tan−1⁡815 then

α = sin − 1 ⁡ 36 85 , β = cos − 1 ⁡ 4 5 and γ = tan − 1 ⁡ 8 15 sin α = 36 85 cos β = 4 5 tan γ = 8 15 ⇒ tan α = 77 85 , tan β = 3 4 tan ( β + γ ) = tan β + tan γ 1 - tan β tan γ = 3 4 + 8 15 1 - 3 4 8 15 = 45 + 32 60 - 24 = - 77 36 = - tan γ tan β + tan γ 1 - tan β tan γ = - tan α ⇒ tan α + tan β + tan γ = tan α tan β tan γ c o t α c o t β + c o t β c o t γ + c o t γ c o t α = 1

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Let $α = \sin^{- 1} (\frac{36}{85}), β = \cos^{- 1} (\frac{4}{5}) and γ = \tan^{- 1} (\frac{8}{15})$ then

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$\cot α + \cot β + \cot γ = \cot αcot βcot γ$

$\tan α + \tan β + \tan γ = \tan αtan βtan γ$

$\cot αcot β + \cot βcot γ + \cot αcot γ = 1$

$\tan αtan β + \tan βtan γ + \tan αtan γ = 1$

answer is A, B.

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Detailed Solution

$α = \sin^{- 1} (\frac{36}{85}), β = \cos^{- 1} (\frac{4}{5}) and γ = \tan^{- 1} (\frac{8}{15})$
$\sin α = \frac{36}{85} \cos β = \frac{4}{5} \tan γ = \frac{8}{15} \Rightarrow \tan α = \frac{77}{85}, \tan β = \frac{3}{4}$
$\begin{array}{rcl} \tan (β + γ) & = & \frac{\tan β + \tan γ}{1 - \tan β \tan γ} \\ = & \frac{\frac{3}{4} + \frac{8}{15}}{1 - \frac{3}{4} \frac{8}{15}} \\ = & \frac{45 + 32}{60 - 24} \\ = & - \frac{77}{36} \\ = & - \tan γ \end{array}$
$\frac{\tan β + \tan γ}{1 - \tan β \tan γ} = - \tan α \Rightarrow \tan α + \tan β + \tan γ = \tan α \tan β \tan γ$
$c o t α c o t β + c o t β c o t γ + c o t γ c o t α = 1$