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प्रश्न
Prove that `(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A . "cosec" A + 1`.
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उत्तर
`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`
= `((cos A)/(sin A))/(1 - (sin A)/(cos A)) + ((sin A)/(cos A))/(1 - (cos A)/(sin A))`
= `((cos A)/(sin A))/((cos A - sin A)/(cos A)) + ((sin A)/(cos A))/((sin A - cos A)/(sin A))`
= `(cos A)/(sin A) xx (cos A)/(cos A - sin A) + (sin A)/(cos A) xx (sin A)/(sin A - cos A)`
= `(cos^2A)/(sin A(cos A - sin A)) + (sin^2A)/(cos A(sin A - cos A))`
= `1/(sin A - cos A) ((-cos^3A + sin^3A)/(sin A cos A))`
= `1/(sin A - cos A)((sin^3A - cos^3A)/(sin A cos A))`
= `1/(sin A - cos A) xx ((sin A - cos A)(sin^2A + sin A cos A + cos^2A))/(sin A cos A)` ...[∵ a3 – b3 = (a – b)(a2 + ab + b2)]
= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)` ...(i)
= `(1 + sin A cos A)/(sin A cos A)` ...[∵ sin2A + cos2A = 1]
= `1/(sin A cos A) + (sin A cos A)/(sin A cos A)`
= cosec A sec A + 1 ...(ii)
`(cot A)/(1 - tan A) + (tan A)/(1 - cot A)`
= `(sin^2A + sin A cos A + cos^2A)/(sin A cos A)` ...[From (i)]
= `(sin^2A)/(sin A cos A) + (sin A cos A)/(sin A cos A) + (cos^2A)/(sin A cos A)`
= `(sin A)/(cos A) + 1 + (cos A)/(sin A)`
= tan A + 1 + cot A ...(iii)
From (ii) and (iii), we get
`(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A . "cosec" A + 1`
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