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प्रश्न
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
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उत्तर
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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= `cos^2theta xx square .....[1 + tan^2theta = square]`
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= 12
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In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
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