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प्रश्न
Prove the following identities:
`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
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उत्तर १
`cotA/(1 - tanA) + tanA/(1 - cotA)`
= `(1/tanA)/(1 - tanA) + tanA/(1 - 1/tanA)`
= `1/(tanA(1 - tanA)) + tan^2A/(tanA - 1)`
= `(1 - tan^3A)/(tanA(1 - tanA))`
= `((1 - tanA)(1 + tanA + tan^2A))/(tanA(1 - tanA))`
= `(1 + tanA + tan^2A)/tanA`
= cot A + 1 + tan A
उत्तर २
L.H.S. = `cotA/(1 - tanA) + tanA/(1 - cotA)`
= `((cosA/sinA))/((1/1 - sinA/cosA)) + ((sinA/cosA))/((1/1 - cosA/sinA))`
= `((cosA/sinA))/(((cosA - sinA)/cosA)) + ((sinA/cosA))/(((sinA - cosA)/sinA))`
= `(cos^2A)/(sinA(cosA - sinA)) + (sin^2A)/(cosA(sinA - cosA))`
= `(cos^2A)/(sinA(cosA - sinA)) - (sin^2A)/(cosA(cosA - sinA))`
= `(cos^3A - sin^3A)/(sinAcosA(cosA - sinA))`
= `(\cancel((cosA - sinA))(cos^2A + cosAsinA + sin^2A))/(sinAcosA\cancel((cosA - sinA)))`
= `(cos^2A + cosA sinA + sin^2A)/(sinAcosA)`
= `(\cancel(cos^2A))/(sinA\cancel(cosA)) + (\cancel(cosAsinA))/(\cancel(sinAcosA)) + (\cancel(sin^2A))/(\cancel(sinA)cosA)`
= cos A + 1 + tan A
= 1 + tan A + cot A
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