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Question
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
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Solution 1
L.H.S. = `1/(secA + tanA)`
= `1/(1/cosA + sinA/cosA)`
= `1/((1 + sinA)/cosA)`
= `cosA/(1 + sinA) xx (1 - sinA)/(1 + sinA)`
= `(cosA(1 - sinA))/((1)^2 - sin^2A)`
= `(cosA(1 - sinA))/cos^2A`
= `1/cosA - sinA/cosA`
= sec A – tan A
L.H.S. = R.H.S.
Hence proved.
Solution 2
L.H.S = `1/(secA + tanA)`
= `((secA - tanA))/((secA + tanA)(secA - tanA))` ...((Multiply Num. and Deno. by sec A – tan A)
= `(secA - tanA)/(sec^2A - tan^2A)`
= `(secA - tanA)/1` ...[∵ sec2 A – tan2 A = 1]
= sec A – tan 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`
