If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2θ
Let ΔABC in which ∠B = 90º and ∠C = θ
cot θ = BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2
AC2 = (8k)2 + (7k)2
AC2 = 64k2 + 49k2
AC2 = 113k2
AC = √113 k
