Advertisements
Advertisements
प्रश्न
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
Advertisements
उत्तर
L.H.S. = `((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA))`
= `(cosec^2A + cot^2A - 2cosecAcotA + 1)/(secA(cosecA - cotA))`
= `(cosec^2A + (1 + cot^2A) - 2cosecAcotA)/(secA(cosecA - cotA))`
= `(cosec^2A + cosec^2A - 2cosecAcotA)/(secA(cosecA - cotA))`
= `(2cosec^2A - 2cosecAcotA)/(secA(cosecA - cotA))`
= `(2cosecA(cosecA - cotA))/(secA(cosecA - cotA))`
= `(2cosecA)/secA`
= `(2 1/sinA)/(1/cosA)`
= `2/sinA xx cosA/1`
= `2 cosA/sinA`
= 2 cot A = R.H.S.
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
If \[sec\theta + tan\theta = x\] then \[tan\theta =\]
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Prove the following identity :
`2(sin^6θ + cos^6θ) - 3(sin^4θ + cos^4θ) + 1 = 0`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ prove that x2 - y2 = a2 - b2.
Prove that sin( 90° - θ ) sin θ cot θ = cos2θ.
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
