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

2 Answers
Ratnaker Mehta
Jul 19, 2016

See the proof given below.

Explanation:

Let #secalpha=a, secbeta=b and secgamma=c#.

Since, #AA theta in RR-{(2k+1)pi/2| k in ZZ}, |sectheta| >=1#, we find

that, #|a|>=1, |b|>=1, |c|>=1#.

#rArr tanalpha=+-sqrt(a^2-1), tanbeta=+-sqrt(b^2-1)#, and,

#tangamma=+-sqrt(c^2-1)# are defined as #|a|, |b|, |c| >=1#.

Then, by what is given, we have,

#a=bc+-sqrt((b^2-1)(c^2-1)) rArr(a-bc)^2=(b^2-1)(c^2-1)#

#rArr a^2-2abc+cancel(b^2c^2)=cancel(b^2c^2)-b^2-c^2+1#

#rArr b^2-2abc=1-a^2-c^2#

Adding #a^2c^2# on both sides, we have,

#b^2-2abc+a^2c^2=a^2c^2-a^2-c^2+1#

#rArr (b-ac)^2=a^2(c^2-1)-1(c^2-1)=(a^2-1)(c^2-1)#

#rArr b-ac=+-sqrt((a^2-1)(c^2-1))#

#rArr b=ac+-sqrt((a^2-1)(c^2-1)#

#rArr secbeta=secalphasecgamma+-tanalphatangamma#.

Hence, the Proof. Enjoy Maths.!

P dilip_k
Jul 19, 2016

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)#

Shown

Related questions
See all questions in Proving Identities
Impact of this question
1679 views around the world
You can reuse this answer
Creative Commons License