Given $secalpha=secbetasecgamma+tanbetatangamma$ How will you show? $secbeta =secgammasecalpha+-tangammatanalpha$

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Answer 1 · 2016-07-02T01:49:51

As below

Given relation

$secalpha=secbetasecgamma+tanbetatangamma$

$=>secalpha-secbetasecgamma=tanbetatangamma$

$color(green)("Squaring both sides")$

$=>(secalpha-secbetasecgamma)^2=tan^2betatan^2gamma$

$=>sec^2alpha+sec^2betasec^2gamma-2secalphasecbetasecgamma=tan^2betatan^2gamma$

$=>-2secalphasecbetasecgamma=-sec^2alpha+tan^2betatan^2gamma-sec^2betasec^2gamma$

$=>-2secalphasecbetasecgamma=-sec^2alpha+(sec^2beta-1)(sec^2gamma-1) -sec^2betasec^2gamma$

$=>-2secalphasecbetasecgamma=-sec^2alpha+cancel(sec^2betasec^2gamma)+1-sec^2gamma-sec^2beta-cancel(sec^2betasec^2gamma)$

$=>sec^2beta-2secalphasecbetasecgamma=-sec^2alpha+1-sec^2gamma$

$color(blue)("Adding "(sec^2gammasec^2alpha)" both sides "$

$=>sec^2beta-2secalphasecbetasecgamma+sec^2gammasec^2alpha=sec^2gammasec^2alpha-sec^2alpha+1-sec^2gamma$

$=>(secbeta-secgammasecalpha)^2=sec^2alpha(sec^2gamma-1)-1(sec^2gamma-1)$

$=>(secbeta-secgammasecalpha)^2=(sec^2gamma-1)(sec^2alpha-1)$

$=>(secbeta-secgammasecalpha)^2=tan^2gammatan^2alpha$

$=>(secbeta-secgammasecalpha)=+-sqrt(tan^2gammatan^2alpha)$

$=>secbeta-secgammasecalpha=+-tangammatanalpha$

$=>color(BLUE)(secbeta=secgammasecalpha+-tangammatanalpha)$

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Given secalpha=secbetasecgamma+tanbetatangammasecalpha=secbetasecgamma+tanbetatangamma How will you show? secbeta =secgammasecalpha+-tangammatanalphasecbeta =secgammasecalpha+-tangammatanalpha