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प्रश्न
`(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"` हे सिद्ध करा.
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उत्तर
डावी बाजू = `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1)`
= `(cot"A" + "cosec A" - ("cosec"^2"A" - cot^2"A"))/(cot"A" - "cosec A" + 1)` .....`[(because 1 + cot^2"A" = "cosec"^2"A"),(therefore "cosec"^2"A" - cot^2"A" = 1)]`
= `(cot"A" + "cosec A" - ("cosec A" + cot"A")("cosec A" - cot"A"))/(cot"A" - "cosec A" + 1)` .....[∵ a2 – b2 = (a + b) (a – b)]
= `((cot"A" + "cosec A")(1 - "cosec A" + cot "A"))/(cot"A" - "cosec A" + 1)`
= cot A + cosec A
= `"cos A"/"sin A" + 1/"sin A"`
= `(cos "A" + 1)/"sin A"`
= उजवी बाजू
∴ `(cot "A" + "cosec A" - 1)/(cot"A" - "cosec A" + 1) = (1 + cos "A")/"sin A"`
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संबंधित प्रश्न
(sec θ + tan θ) (1 - sin θ) = cos θ
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sin6A + cos6A = 1 – 3sin2A . cos2A हे सिद्ध करा.
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