Advertisements
Advertisements
प्रश्न
Prove that:
`1/(sinA - cosA) - 1/(sinA + cosA) = (2cosA)/(2sin^2A - 1)`
Advertisements
उत्तर
`1/(sinA - cosA) - 1/(sinA + cosA)`
= `(sinA + cosA - sinA + cosA)/((sinA - cosA)(sinA + cosA)`
= `(2cosA)/(sin^2A - cos^2A)`
= `(2cosA)/(sin^2A - (1 - sin^2A))`
= `(2cosA)/(sin^2A - 1 + sin^2A)`
= `(2cosA)/(2sin^2A - 1)`
संबंधित प्रश्न
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Prove that sec2θ – cos2θ = tan2θ + sin2θ
If 2sin2β − cos2β = 2, then β is ______.
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
