Question

If cot θ = 7/8 evaluate `((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ))`

Answer 1

Let us consider a right triangle ABC, right-angled at point B.

`cot theta = 7/8`

If BC is 7k, then AB will be 8k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we obtain

AC² = AB² + BC²

= (8k)² + (7k)²

= 64k² + 49k²

= 113k²

`AC = sqrt113k`

`sin theta =  (8k)/sqrt(113k) =  8/sqrt(113)`

`cos theta =  (7k)/sqrt(113k) = 7/sqrt113`

`((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ)) = (1-sin^2θ)/(1-cos^2θ)`

`= (1-(8/sqrt113)^2)/(1-(7/sqrt(113))^2)= (1-64/113) /(1-49/113)`

`= (49/113)/(64/113) = 49/64`

Answer 2

`cot theta = 7/8`

`((1+sin θ )(1-sin θ))/((1+cos θ)(1-cos θ))`

`= (1 - sin^2 theta)/(1 - cos^2 theta)`   [∵ (a + b) (a – b) = a² − b²] a = 1, b = sin 𝜃

We know that sin 2𝜃 + cos2 𝜃 = 1

1 − sin² 𝜃 = cos² 𝜃 = cos² 𝜃

1 − cos² 𝜃 = sin² 𝜃

`= (cos^2 theta)/(sin^2 theta)`

`= cot^2 theta`

`= (cot theta)^2 = [7/8]^2`

`= 49/64`