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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
संबंधित प्रश्न
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
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`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
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`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
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`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
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`sqrt((1 - sinA)/(1 + sinA)) = cosA/(1 + sinA)`
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
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