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प्रश्न
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
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उत्तर १
`1/(cosA + sinA) + 1/(cosA - sinA)`
= `(cosA + sinA + cosA - sinA)/((cosA + sinA)(cosA - sinA)`
= `(2cosA)/(cos^2A - sin^2A)`
= `(2cosA)/(cos^2A - (1 - cos^2A))`
= `(2cosA)/(cos^2A - 1 + cos^2A)`
= `(2cosA)/(2cos^2A - 1)`
उत्तर २
`1/(cosA + sinA) + 1/(cosA - sinA)`
`1/(cosA + sinA) + 1/(cosA - sinA) = ((cos A - sin A) + (cos A + sin A))/((cos A + ain A)(cos A - sin A))`
(cosA − sinA) + (cosA + sinA) = 2 cosA
(cosA + sinA) (cosA − sinA) = cos2A − sin2A = cos(2A)
cos(2A) = 2cos2A − 1
`(2cosA)/cos(2A) = (2cosA)/(2cos^2 A-1)`
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
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