Advertisements
Advertisements
प्रश्न
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
Advertisements
उत्तर
(tan A + cot A) (cosec A – sin A) (sec A – cos A)
= `(sinA/cosA + cosA/sinA)(1/sinA - sinA)(1/cosA - cosA)`
= `((sin^2A + cos^2A)/(sinAcosA))((1 - sin^2A)/sinA)((1 - cos^2A)/cosA)`
= `(1/(sinAcosA))(cos^2A/sinA)(sin^2A/cosA)`
= 1
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(1+ secA)/sec A = (sin^2A)/(1-cosA)`
[Hint : Simplify LHS and RHS separately.]
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities.
`tan A/(1 + tan^2 A)^2 + cot A/((1 + cot^2 A)) = sin A cos A`
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
Prove the following identities:
(cos A + sin A)2 + (cos A – sin A)2 = 2
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove that: 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0.
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
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 `tan θ = 7/24`, then to find value of cos θ complete the activity given below.
Activity:
sec2θ = 1 + `square` ...[Fundamental tri. identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square/576`
sec2θ = `square/576`
sec θ = `square`
cos θ = `square` ...`[cos theta = 1/sectheta]`
