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प्रश्न
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
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उत्तर
`(p^2 - 1)/(p^2 + 1)`
= `((secA + tanA)^2 - 1)/((secA + tanA)^2 + 1)`
= `(sec^2A + tan^2A + 2tanA secA - 1)/(sec^2A + tan^2A + 2tanA secA + 1)`
= `(tan^2A + tan^2A + 2tanA secA)/(sec^2A + sec^2A + 2tanA secA)`
= `(2tan^2A + 2tanA secA)/(2sec^2A + 2tanA secA)`
= `(2tanA(tanA + secA))/(2secA(tanA + secA)`
= sin A
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संबंधित प्रश्न
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
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`cosA/(1 - sinA) = sec A + tan A`
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`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
(sec θ + tan θ) . (sec θ – tan θ) = ?
Prove that `(1 + sec theta - tan theta)/(1 + sec theta + tan theta) = (1 - sin theta)/cos theta`
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
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
