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Question
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
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Solution
L.H.S. = `1/(cosA + sinA - 1) + 1/(cosA + sinA + 1)`
= `(cosA + sinA + 1 + cosA + sinA - 1)/((cosA + sinA)^2 - 1)`
= `(2(cosA + sinA))/(cos^2A + sin^2A + 2cosAsinA - 1)`
= `(2(cosA + sinA))/(1 + 2cosAsinA - 1)`
= `(cosA + sinA)/(cosAsinA)`
= `cosA/(cosAsinA) + sinA/(cosAsinA)`
= `1/sinA + 1/cosA`
= cosec A + sec A = R.H.S.
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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`
