Let secalpha=a, secbeta=b and secgamma=csecα=a,secβ=bandsecγ=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.!
