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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]
