Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Advertisements
उत्तर
LHS = `(cosecA - sinA)(secA - cosA)`
= `(1/sinA - sinA)(1/cosA - cosA)`
= `((1-sin^2A)/(sinA))((1 - cos^2A)/cosA)`
= `(cos^2A/sinA)(sin^2A/cosA)` = cosA.sinA
RHS = `1/(tanA + cotA)`
= `1/(sinA/cosA + cosA/sinA) = 1/((sin^2A + cos^2A)/(sinA.cosA))` = cosA.sinA
Hence , LHS = RHS
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A`
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that `x^2/a^2 + y^2/b^2 - x^2/c^2 = 1`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
If 2sin2θ – cos2θ = 2, then find the value of θ.
