Question #b4d99
1 Answer
Feb 24, 2018
#x=a secA sinB#
#=>(x/a)^2=sec^2A cdot sin^2B" "...(1)#
#y=b secA cosB#
#=>(y/b)^2=sec^2A cdot cos^2B" "...(2)#
#z=c tanA#
#=>(z/c)^2=tan^2A" "...(3)# From, 1st, 2nd and 3rd equation, we get
#(x/a)^2+(y/b)^2+(z/c)^2#
#=sec^2A cdot sin^2B+sec^2A cdot cos^2B+tan^2A#
#=sec^2A(sin^2B+cos^2B)+tan^2A#
#=sec^2A+tan^2A#
