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Question
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
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Solution
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.
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Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
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Prove that `(1 + sec "A")/"sec A" = (sin^2"A")/(1 - cos"A")`
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
