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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
9 sec2 A − 9 tan2 A = ______.
Prove the following trigonometric identities.
`(cos^2 theta)/sin theta - cosec theta + sin theta = 0`
Prove the following identities:
`1/(secA + tanA) = secA - tanA`
Prove the following identities:
`(sinA - cosA + 1)/(sinA + cosA - 1) = cosA/(1 - sinA)`
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
Prove that `(cot A)/(1 - cot A) + (tan A)/(1 - tan A) = -1`.
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
