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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