Advertisements
Advertisements
Question
Show that `sqrt((1+cosA)/(1-cosA)) = cosec A + cot A`
Advertisements
Solution
L.H.S = `sqrt((1-cosA)/(1+cos A))`
`= sqrt((1-cosA)/(1+cosA) xx (1 - cos A)/(1- cos A)) = sqrt((1- cosA)^2/(1-cos^2A))`
`=sqrt((1- cosA)^2/(sin^2A)) = (1-cosA)/sin A = 1/sin A - cos A/sin A = cosec A -cot A` = R.H.S
Hence prove.
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove that:
(sin A + cos A) (sec A + cosec A) = 2 + sec A cosec A
`1 + (tan^2 θ)/((1 + sec θ)) = sec θ`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
If `sec theta + tan theta = x," find the value of " sec theta`
If sin θ = `11/61`, find the values of cos θ using trigonometric identity.
9 sec2 A − 9 tan2 A is equal to
Prove the following identity :
`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ`
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
If `sin θ + cos θ = sqrt(3)`, then show that tan θ + cot θ = 1.
If tan θ – sin2θ = cos2θ, then show that `sin^2θ = 1/2`.
If cos 9α = sinα and 9α < 90°, then the value of tan5α is ______.
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
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 that `(cot A - cos A)/(cot A + cos A) = (cos^2 A)/(1 + sin A)^2`
