Question

Prove that sec72- sec36- sec60 =0. Fast its urgent

Answer 1

heya

sec36°=1/√5+1/4.

sec72°=1/√5-1 /4


and ,sec60°=2/√3
then ,putting on sec72°-sec36°-sec60°

=)4/√5-1-4/√5+1-√3/2

=)4(√5+1)-4(√5-1)/4=√3/2

=)8=√3/2

I think your question will be wrong.

Answer 2

sec72- sec36- sec60 =0
so we can frame the eqn,
sec72- sec36 = sec60 
sec72-sec36 = $\frac{1}{cos72} -\frac{1}{cos36}$
                       =$\frac{cos36-cos72}{cos36.cos72}$
                       =$\frac{2sin( \frac{72+36}{2} ).sin(\frac{72-36}{2} )}{sin54.sin18} ( 40) (1) and (2)$
                       =$\frac{2sin54.sin18}{sin54.sin18}$
                       =2
                       =sec 60
hence,
LHS=RHS