InterviewSolution
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InterviewSolution
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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`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` |
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Answer» Correct Answer - A::B `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` |
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