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प्रश्न
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
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उत्तर
L.H.S. = `(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2`
= `(sinA/cosA + 1/cosA)^2 + (sinA/cosA - 1/cosA)^2`
= `((sinA + 1)/cosA)^2 + ((sinA - 1)/cosA)^2`
= `(sinA + 1)^2/(cos^2A) + (sinA - 1)^2/(cos^2A)`
= `((sinA + 1)^2 + (sinA - 1)^2)/(cos^2A)`
= `(sin^2A + 1 + 2sinA + sin^2A + 1 - 2sinA)/cos^2A`
= `(2sin^2A + 2)/(cos^2A)`
= `(2(1 + sin^2A))/(1 - sin^2A)`
= `2((1 + sin^2A)/(1 - sin^2A))` = R.H.S.
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