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Question
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
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Solution 1
L.H.S = `(tan^2 theta)/(sec theta - 1)^2 `
`= (sec^2 theta - 1)/(sec theta - 1)^2`
`= ((sec theta - 1)(sec theta + 1))/(sec theta - 1)^2`
`= (sec theta + 1)/(sec theta - 1)`
`= (1/cos theta + 1)/(1/cos theta - 1)`
`= ((1 + cos theta)/cos theta)/((1 - cos theta)/cos theta)`
`= (1 + cos theta)/(1 - cos theta)`
= R.H.S
Solution 2
L.H.S = `(tan^2 θ)/(sec θ - 1)^2 `
= `(sin^2 θ/cos^2 θ)/(1/cos θ - 1)^2 .....( ∵ tan θ = sin θ /cos θ )`
= `(sin^2 θ/cos^2 θ)/((1/cos θ - 1)^2/(cos^2 θ)) ....( ∵ sec θ = 1/cos θ) `
= `(sin^2 θ)/( 1 - cos θ)^2 ....( ∵ sin^2 θ = 1 - cos^2 θ) `
= `( 1 - cos^2 θ)/( 1 - cos θ)^2`
= `( 1 - cos θ)( 1 + cos θ)/( 1 - cos θ)^2` ....( ∵ a2 - b2 = (a + b)(a - b))
= `( 1 + cos θ)/( 1 - cos θ)`
= RHS.
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
