Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
Advertisements
उत्तर
LHS = `(1 + cotA + tanA)(sinA - cosA)`
= `(1 + cosA/sinA + sinA/cosA)(sinA - cosA)`
= `((sinAcosA + cos^2A + sin^2A)/(sinAcosA))(sinA - cosA)`
= `((sin^3A - cos^3A))/(sinAcosA)` (∵(`sin^3A - cos^3A) = (sinA - cosA)(sinA cosA + cos^2A + sin^2A`))
= `sin^3A/(sinAcosA) - cos^3A/(sinAcosA)`
= `sin^2A/cosA - cos^2A/sinA = 1/cosA xx sin^2A - 1/sinA xx cos^2A`
= `secAsin^2A - cosecAcos^2A`
= `secA/(cosec^2A) - (cosecA)/sec^2A`
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
`sec theta (1- sin theta )( sec theta + tan theta )=1`
Show that none of the following is an identity:
`sin^2 theta + sin theta = 2`
What is the value of 9cot2 θ − 9cosec2 θ?
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
