Advertisements
Advertisements
प्रश्न
\[\frac{x^2 - 1}{2x}\] is equal to
विकल्प
sec θ + tan θ
sec θ − tan θ
sec2 θ + tan2 θ
sec2 θ − tan2 θ
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:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
Prove that:
(sec A − tan A)2 (1 + sin A) = (1 − sin A)
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
Prove that:
`cot^2A/(cosecA - 1) - 1 = cosecA`
`(tan theta)/((sec theta -1))+(tan theta)/((sec theta +1)) = 2 sec theta`
Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.
Prove the following identity :
tanA+cotA=secAcosecA
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`
Prove that `sqrt((1 + sin θ)/(1 - sin θ))` = sec θ + tan θ.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
If sin θ + cos θ = `sqrt(3)`, then prove that tan θ + cot θ = 1
If 1 – cos2θ = `1/4`, then θ = ?
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ
Prove that sec2θ – cos2θ = tan2θ + sin2θ
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B
