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Question
Prove that `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot θ`.
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Solution
L.H.S. = `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1)`
= `(sin θ)/(1/cos θ + 1) + (sin θ)/(1/(cos θ) - 1`
= `(sin θ)/((1 + cos θ)/(cos θ)) + (sin θ)/((1 - cos θ)/(cos θ))`
= `(sin θ cos θ)/(1 + cos θ) + (sin θ cos θ)/(1 - cos θ)`
= `sin θ cos θ (1 /(1 + cos θ) + 1/(1 - cos θ))`
= `sin θ cos θ [(1 - cos θ + 1 + cos θ)/((1 + cos θ)(1 - cos θ))]`
= `sin θ cos θ (2/(1 - cos^2θ))` ...[∵ (a + b)(a – b) = a2 – b2]
= `sin θ cos θ xx 2/(sin^2θ)` ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`
= `2 xx (cos θ)/(sin θ)`
= 2 cot θ
= R.H.S.
∴ `(sin θ)/(sec θ + 1) + (sin θ)/(sec θ - 1) = 2 cot 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`
