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प्रश्न
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
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उत्तर
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2`
= `(sinθ/cosθ + 1/cosθ)^2 + (sinθ/cosθ - 1/cosθ)^2`
= `((sinθ + 1)/cosθ)^2 + ((sinθ - 1)/cosθ)^2`
= `(sinθ + 1)^2/(cos^2θ) + (sinθ - 1)^2/cos^2θ`
= `((sinθ + 1)^2 + (sinθ - 1)^2)/cos^2A`
= `(sin^2θ + 1 + 2sinθ + sin^2θ + 1 - 2sinθ)/(1 - sin^2θ)`
= `(2(1 + sin^2θ))/(1 - sin^2θ)`
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संबंधित प्रश्न
Prove the following trigonometric identities:
`(1 - cos^2 A) cosec^2 A = 1`
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`cotA/(1 - tanA) + tanA/(1 - cotA) = 1 + tanA + cotA`
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
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`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
cos θ . sec θ = ?
Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.
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
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
