Let `alpha=som^(-1)((36)/(85)),beta=cos^(-1)(4/5)a n dgamma=tan^(-1)(8/(15))` then `cotalpha+cotbeta+cotgamma=cotalphacotbetacotgamma` `tanalphatanbeta+tanbetatangamma+tanalphatangamma=1` `tanalpha+tanbeta+tangamma=tanalphatanbetatangamma` `cotalphacotbeta+cotbetacotgamma+cotalphacotgamma=1`

Options

A. `cot alpha + cot beta + cot gamma = cot alpha cot beta cot gamma`
B. `tan alpha tan beta + tan beta tan gamma + tan alpha tan gamma = 1`
C. `tan alpha + tan beta + tan gamma = tan alpha tan beta tan gamma`
D. `cot alpha cot beta + cot beta cot gamma + cot alpha cot gamma = 1`

Answer

A, B

Explanation

`alpha = sin^(-1).(36)/(85) rArr alpha = (36)/(85) rArr tan alpha = (36)/(77)`
`beta = cos^(-1) ((4)/(5)) rArr cos beta = (4)/(5) rArr tan beta = (3)/(4)`
and `tan gamma = (8)/(15)`
`:. tan (alpha + beta + gamma) = (tan alpha + tan beta + tan gamma - tan alpha tan beta tan gamma)/(1-tan alpha tan beta - tan beta tan gamma - tan gamma tan alpha)`
`= ((36)/(77) + (3)/(4) + (8)/(15) -(36)/(77) .(3)/(4).(8)/(15))/(1-((36)/(77) xx (3)/(4) + (8)/(15) xx (3)/(4) + (8)/(15) xx (36)/(77)))`
`rArr tan (alpha + beta + gamma) = oo`
`rArr alpha + beta + gamma =(pi)/(2)`
`rArr cot alpha + cot beta + cot gamma = cot alpha cot beeta cot gamma`
`rArr tan alpha tan beta + tan beta tan gamma + tan alpha tan gamma = 1`