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प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
(cosec A - sin A) (sec A - cos A) = `1/(tanA+cotA)`
[Hint: Simplify LHS and RHS separately.]
उत्तर
(cosec A – sin A) (sec A – cos A) = `1/(tanA+cotA)`
L.H.S. = (cosec A – sin A) (sec A – cos A)
= `(1/sinA-sinA)(1/cosA-cosA)`
= `((1-sin^2A)/sinA)((1-cos^2A)/cosA)`
= `((cos^2A)(sin^2A))/(sinAcosA)`
= sinA cosA
R.H.S = `1/(tanA+cotA)`
= `1/(sinA/cosA+cosA/sinA)`
= `1/((sin^2A + cos^2A)/(sinAcosA))`
= `(sinAcosA)/(sin^2A+cos^2A)`
= sinA cosA
Hence, L.H.S = R.H.S
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