Advertisements
Advertisements
प्रश्न
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
उत्तर
LHS = sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ )
= sec θ. sec θ - tan θ. tan θ
= sec2θ - tan2θ
= 1
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following trigonometric identities.
`(tan^3 theta)/(1 + tan^2 theta) + (cot^3 theta)/(1 + cot^2 theta) = sec theta cosec theta - 2 sin theta cos theta`
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
`tan theta /((1 - cot theta )) + cot theta /((1 - tan theta)) = (1+ sec theta cosec theta)`
If cosθ = `5/13`, then find sinθ.
Prove that `sqrt((1 + cos "A")/(1 - cos"A"))` = cosec A + cot A
If 1 + sin2θ = 3sinθ cosθ, then prove that tanθ = 1 or `1/2`.
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
