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प्रश्न
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
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उत्तर
L.H.S. = `(secA - tanA)/(secA + tanA)`
= `(secA - tanA)/(secA + tanA) xx (secA - tanA)/(secA - tanA)`
= `(secA - tanA)^2/(sec^2A - tan^2A)`
= `(sec^2A + tan^2A - 2secAtanA)/1`
= 1 + tan2 A + tan2 A – 2 sec A tan A
= 1 – 2 sec A tan A + 2 tan2 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`
