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Question
Prove the following identities.
`costheta/(1 + sintheta)` = sec θ – tan θ
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Solution
`cos theta/(1 + sin theta)` = sec θ – tan θ
R.H.S. = sec θ – tan θ
= `1/cos theta - sin theta/cos theta`
= `(1 - sin theta)/costheta`
= `(1 - sin theta)/cos theta xx (1 + sin theta)/(1 + sin theta)`
= `(1 - sin^2 theta)/(cos theta(1 + sin theta)`
= `cos^2 theta/(cos theta(1 + sin theta))`
= `costheta/(1 + sintheta)`
L.H.S. = R.H.S.
∴ `cos theta/(1 + sin theta)` = sec θ – tan θ
Aliter:
L.H.S. = `cos theta/(1 - sin theta)` ...[conjugate (1 – sin θ)]
= `(cos theta(1 + sin theta))/((1 - sin theta)(1 + sin theta))`
= `(cos theta(1 + sin theta))/((1 - sin^2 theta))`
= `(cos theta (1 + sin theta))/(cos^2 theta)`
= `(1 + sin theta)/costheta`
L.H.S. = 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`
