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 trigonometric identities.
`sin theta/(1 - cos theta) = cosec theta + cot theta`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
`sin theta/((cot theta + cosec theta)) - sin theta /( (cot theta - cosec theta)) =2`
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
Prove that: `1/(cosec"A" - cot"A") - 1/sin"A" = 1/sin"A" - 1/(cosec"A" + cot"A")`
Prove that `(sec A)/(tan A + cot A) = sin A`.
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
`(cos^2 θ)/(sin^2 θ) - 1/(sin^2 θ)`, in simplified form, is ______.
