हिंदी

Prove the following identities, where the angles involved are acute angles for which the expressions are defined: 1+secAsecA=sin2A1-cosA [Hint : Simplify LHS and RHS separately.]

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प्रश्न

 
 

Prove the following identities, where the angles involved are acute angles for which the expressions are defined:

`(1+ secA)/sec A = (sin^2A)/(1-cosA)` 

[Hint : Simplify LHS and RHS separately.]

 
 
योग
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उत्तर

 

 L.H.S

`(1+secA)/secA = (1+1/(cosA))/(1/cosA)`

= `((cosA+1)/cosA)/(1/cosA)`

= `(cosA+1)`

= `((1-cosA)(1+cosA))/(1-cosA)`

= `(1-cos^2A)/(1-cosA)`

= `(sin^2A)/(1-cosA)`           ...[∵ 1cos2 A = sin2A]

R.H.S

 
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अध्याय 8: Introduction to Trigonometry - EXERCISE 8.3 [पृष्ठ १३१]

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एनसीईआरटी Mathematics [English] Class 10
अध्याय 8 Introduction to Trigonometry
EXERCISE 8.3 | Q 4. (iv) | पृष्ठ १३१

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