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Question
Prove that `(sin θ tan θ)/(1 - cos θ) = 1 + sec θ.`
Solution
LHS = `( sin θ tan θ)/(1 - cos θ)`
= `(sin θ. (sin θ)/(cos θ))/(1 - cos θ)`
= `sin^2 θ/(cos θ( 1 - cos θ))`
= `((1 - cos θ)(1 + cos θ))/(cos θ(1 - cos θ))`
= `(1 + cos θ)/(cos θ) = 1/(cos θ) + cos θ/cos θ`
= sec θ + 1
= RHS
Hence proved.
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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`
