Advertisements
Advertisements
प्रश्न
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
Advertisements
उत्तर
L.H.S. = `(1 + sec A)/sec A`
= `(1 + 1/ cos A)/(1/cos A)`
= `((cos A + 1)/cos A)/(1/cos A)`
= 1 + cos A = `((1 + cos A))/1 xx ((1 - cos A))/((1 - cos A))`
= `(1 - cos^2 A)/(1 - cos A)`
`\implies (sin^2 A)/(1 - cos A)` = R.H.S. ...(∵ sin2 A + cos2 A = 1)
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
`(1 + cot^2 theta ) sin^2 theta =1`
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos ^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
`{1/((sec^2 theta- cos^2 theta))+ 1/((cosec^2 theta - sin^2 theta))} ( sin^2 theta cos^2 theta) = (1- sin^2 theta cos ^2 theta)/(2+ sin^2 theta cos^2 theta)`
\[\frac{\sin \theta}{1 + \cos \theta}\]is equal to
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
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)
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
