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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) = 1tanA+cotA - Mathematics

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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.] 

योग

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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अध्याय 8: Introduction to Trigonometry - Exercise 8.4 [पृष्ठ १९४]

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एनसीईआरटी Mathematics [English] Class 10
अध्याय 8 Introduction to Trigonometry
Exercise 8.4 | Q 5.09 | पृष्ठ १९४

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