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Question
Prove that cosec θ – cot θ = `(sin θ)/(1 + cos θ)`.
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Solution
L.H.S. = cosec θ – cot θ
= `1/(sin θ) - (cos θ)/(sin θ)`
= `(1 - cos θ)/(sin θ)`
= `(1 - cos θ)/(sin θ) xx (1 + cos θ)/(1 + cos θ)` ...[On rationalising the numerator]
= `(1 - cos^2θ)/(sinθ(1 + cosθ))`
= `(sin^2θ)/(sinθ(1 + cosθ))` ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`
= `(sin θ)/(1 + cos θ)`
= R.H.S.
∴ cosec θ – cot θ = `(sin θ)/(1 + cos θ)`
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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`
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
