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प्रश्न
\[\frac{x^2 - 1}{2x}\] is equal to
पर्याय
sec θ + tan θ
sec θ − tan θ
sec2 θ + tan2 θ
sec2 θ − tan2 θ
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उत्तर
The given expression is `sqrt ((1+sinθ)/(1-sinθ))`
Multiplying both the numerator and denominator under the root by `1+ sinθ`, we have
`sqrt (((1+ sinθ)(1+sin θ))/((1+sin θ)(1-sinθ)))`
`=sqrt((1+sinθ)/((1- sin^2θ)))`
`= sqrt((1+ sinθ)^2/cos^2θ)`
=`(1+sinθ)/cosθ`
=` 1/cosθ+sinθ/cosθ`
=` sec θ+tan θ`
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(sin theta-2sin^3theta)/(2cos^3theta -costheta) = tan theta`
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
`(cot ^theta)/((cosec theta+1)) + ((cosec theta + 1))/cot theta = 2 sec theta`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Without using trigonometric table , evaluate :
`cosec49°cos41° + (tan31°)/(cot59°)`
Find the value of sin 30° + cos 60°.
Prove that `(tan^2"A")/(tan^2 "A"-1) + (cosec^2"A")/(sec^2"A"-cosec^2"A") = (1)/(1-2 co^2 "A")`
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
Choose the correct alternative:
tan (90 – θ) = ?
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Prove that `sec"A"/(tan "A" + cot "A")` = sin A
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
