Question

Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`

Answer 1

L.H.S = `(1 + sec theta - tan theta)/(1 + sec theta + tan theta)`

= `(1 + 1//cos theta - sin theta//cos theta)/(1 + 1//cos theta + sin theta//cos theta)`  ...`[∵ sec theta = 1/cos theta and tan theta = sin theta/cos theta]`

= `(cos theta + 1 - sin theta)/(cos theta + 1 + sin theta)`

= `((cos theta + 1) - sin theta)/((cos theta + 1) + sin theta)`

= `(2 cos^2  theta/2 - 2 sin  theta/2 * cos  theta/2)/(2 cos^2  theta/2 + 2 sin  theta/2 * cos  theta/2)`  ...`[∵ 1 + cos theta = 2 cos^2  theta/2 and sin theta = 2sin  theta/2 cos  theta/2]`

= `(2cos^2  theta/2 - 2 sin  theta/2 * cos  theta/2)/(2cos^2  theta/2 + 2sin  theta/2 * cos  theta/2)`

= `(2cos  theta/2 (cos  theta/2 - sin  theta/2))/(2cos  theta/2(cos  theta/2 + sin  theta/2))`

= `(cos  theta/2 - sin  theta/2)/(cos  theta/2 + sin  theta/2) xx ((cos  theta/2 - sin  theta/2))/((cos  theta/2 - sin  theta/2))`  ...[By rationalisation]

= `(cos  theta/2 - sin  theta/2)^2/((cos^2  theta/2 - sin^2  theta/2))`  ...[∵ (a – b)² = a² + b² – 2ab and (a – b)(a + b) = (a² – b²)]

= `((cos^2  theta/2 + sin^2  theta/2) - (2 sin  theta/2 * cos  theta/2))/cos theta`   ...`[∵ cos^2  theta/2 - sin^2  theta/2 = cos theta]`

= `(1 - sin theta)/cos theta`   ...`[∵ sin^2  theta/2 + cos^2  theta/2 = 1]`

= R.H.S

Hence proved.