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| 1. |
If sec∅+tan∅=x find the value of sec∅ |
| Answer» Given:\xa0sec x + tan x = pAs we know,{tex}sec^2 x - tan^2 x = 1{/tex}{tex}\\therefore{/tex}\xa0{tex}(sec\\ x + tan\\ x)(sec\\ x - tan\\ x) = 1\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}p(sec\\ x - tan\\ x) = 1\xa0{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}sec\\ x - tan\\ x ={/tex}\xa0{tex}\\frac{1}{p}{/tex}Thus, we have{tex}sec\\ x + tan\\ x = p{/tex} and\xa0sec x - tan x =\xa0{tex}\\frac{1}{p}{/tex}{tex}\\Rightarrow{/tex}\xa0(sec x + tan x) + (sec x - tan x) = p +\xa0{tex}\\frac{1}{p}{/tex}\xa0and\xa0(sec x + tan x) - (sec x - tan x) = p -\xa0{tex}\\frac{1}{p}{/tex}{tex}\\Rightarrow{/tex}\xa02 sec x = p +\xa0{tex}\\frac{1}{p}{/tex}\xa0and 2 tan x = -{tex}\\frac{1}{p}{/tex}\xa0{tex}\\Rightarrow{/tex}\xa0sec x =\xa0{tex}\\frac{p^{2}+1}{2 p}{/tex}\xa0and\xa0tan x =\xa0{tex}\\frac{p^{2}-1}{2 p}{/tex} | |