Advertisements
Advertisements
प्रश्न
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
उत्तर
L.H.S. = `sqrt((1 + sin theta)/(1 - sin theta)`
= `sqrt(((1 + sin theta)(1 + sin theta))/((1 - sin theta)(1 + sin theta))` ...[conjugate (1 − sin θ)]
= `sqrt((1 + sin theta)^2/(1 - sin^2 theta)`
= `sqrt((1 + sin theta)^2/(cos^2 theta)`
= `(1 + sin theta)/(cos theta)`
= `1/cos theta + sin theta/cos theta`
= sec θ + tan θ
L.H.S. = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
Prove the following trigonometric identities
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 − y2 = a2 − b2
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Prove the following identities:
`sinA/(1 - cosA) - cotA = cosecA`
The value of sin2 29° + sin2 61° is
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Prove that `tan^3 θ/( 1 + tan^2 θ) + cot^3 θ/(1 + cot^2 θ) = sec θ. cosec θ - 2 sin θ cos θ.`
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
If tan θ = `7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ......[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` .......`[cos theta = 1/sectheta]`
