If cot θ = 7/8, evaluate :
(i) (1 + sin θ)(1 – sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2 θ
Let us assume a △ABC in which ∠B = 90° and ∠C = θ
cot θ = BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positive real number
According to Pythagoras theorem in △ABC we get.
AC2 = AB2 + BC2
AC2 = (8k)2 + (7k)2
AC2 = 64k2 + 49k2
AC2 = 113k2
AC = √113 k
According to the sine and cos function ratios, it is written as
sin θ = AB/AC = Opposite Side/Hypotenuse
= 8k/√113 k = 8/√113 and
cos θ = Adjacent Side/Hypotenuse = BC/AC
= 7k/√113 k = 7/√113
(i) \(\frac{(1+ sin \theta)(1- sin \theta)}{(1+ cos \theta)(1-cos \theta)}\) = \(\frac{1-sin^2 \theta}{1-cos^2 \theta}\)
= \( \frac{1-(\frac{8}{\sqrt{113}})^2}{1-(\frac{7}{\sqrt{113}})^2} \) = \(\frac{49}{64} \)
(ii) \(cot^2 \theta = (\frac{7}{8})^2\) = \(\frac{49}{64}\)