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प्रश्न
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
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उत्तर
L.H.S. = (sin2θ – 1)(tan2θ + 1) + 1
= (– cos2θ) sec2θ + 1
= `- cos^2θ xx 1/(cos^2θ) + 1`
= –1 + 1
= 0
= R.H.S.
Hence Proved.
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संबंधित प्रश्न
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cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
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`sqrt(sec^2A + cosec^2A) = tanA + cotA`
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`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
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