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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
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
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`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove that:
`(sinA - cosA)(1 + tanA + cotA) = secA/(cosec^2A) - (cosecA)/(sec^2A)`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
Prove the following identity :
`(cosecA - sinA)(secA - cosA)(tanA + cotA) = 1`
If sec θ = `25/7`, then find the value of tan θ.
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
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