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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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Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
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= R.H.S.
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