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Question
Prove that `sqrt((1 + cos A)/(1 - cos A)) = (tan A + sin A)/(tan A. sin A)`
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Solution
LHS = `sqrt(((1 + cos A)(1 + cos A))/((1 - cos A)(1 + cos A)))`
= `sqrt((1 + cos A)^2/(1 - cos^2 A))`
= `sqrt((1 + cos^2 A + 2cos A)/sin^2 A`
= `(1 + cos A)/sin A`
RHS = `(tan A + sin A)/(tan A sin A)`
= `(sin A(1/cos A + 1))/((sin A/cos A xx sin A)`
= `(sin A( 1 + cos A))/cos A xx cos A/(sin A sin A)`
= `(1 + cos A)/sin A`
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`
