Advertisements
Advertisements
Question
Prove that:
(cosec A – sin A) (sec A – cos A) sec2 A = tan A
Solution
L.H.S. = (cosec A – sin A) (sec A – cos A) × sec2 A
= `(1/sinA - sinA)(1/cosA - cosA) xx sec^2A` ...`{∵ cosec theta = 1/sintheta, sectheta = 1/costheta, 1 - sin^2theta = cos^2theta, 1 - cos^2theta = sin^2theta}`
= `((1 - sin^2A)/sinA)((1 - cos^2A)/cosA) 1/(cos^2A)`
= `(cos^2A)/(sinA)*(sin^2A)/(cosA)*1/(cos^2A)`
= `sinA/cosA`
= tan A = R.H.S.
APPEARS IN
RELATED QUESTIONS
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a2 + b2 = m2 + n2
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Find the value of sin 30° + cos 60°.
If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`
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 (sec θ + tan θ) (1 – sin θ) = cos θ
