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प्रश्न
Prove the following identities:
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
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उत्तर
R.H.S. = `tan^2A/(secA - 1)^2`
= `(sec^2A - 1)/(secA - 1)^2` ...[sec2θ – tan2θ = 1 sec2θ – 1 = tan2θ]
= `((secA + 1)(secA - 1))/(secA - 1)^2`
= `(secA + 1)/(secA - 1)`
= `(1/(cosA) + 1)/(1/cosA - 1)`
= `((1 + cosA)/cosA)/((1 - cosA)/(cosA))`
= `(1 + cosA)/(1 - cosA)`
R.H.S. = L.H.S.
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Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
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= 12
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