Question

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

If cot θ = `7/8` then find the value of `((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 = 7k, then AB = 8k, where k is a positive integer.

Applying Pythagoras theorem in ΔABC, we get

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)`

= `((113 - 64)/113) /((113 - 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 sin²θ + cos²θ = 1

1 − sin²θ = cos²θ

1 − cos²θ = sin²θ

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

= `cot^2 theta`

= `(cot theta)^2`

= `[7/8]^2`

= `49/64`