Advertisements
Advertisements
Question
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Advertisements
Solution
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.
RELATED QUESTIONS
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
Prove that:
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
`(1+ cos theta + sin theta)/( 1+ cos theta - sin theta )= (1+ sin theta )/(cos theta)`
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
Prove that sin2 5° + sin2 10° .......... + sin2 85° + sin2 90° = `9 1/2`.
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
