Advertisements
Advertisements
Question
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
Advertisements
Solution
LHS = `(sin"A"/cos"A")/(1 - cos"A"/sin"A") + (cos"A"/sin"A")/(1 - sin"A"/cos"A")`
= `(sin"A" sin"A")/(cos"A"(sin"A" - cos"A")) + (cos"A" cos"A")/((cos"A" - sin"A") sin"A"`
= `1/((sin"A" - cos"A")) [(sin^2"A")/cos"A" + (cos^2"A")/(-sin"A")]`
= `(sin^3"A" - cos^3"A")/(sin"A".cos"A"(sin"A" - cos"A"))`
= `((sin"A" - cos"A")(sin^2"A" + cos^2"A" + sin"A". cos"A"))/(sin"A". cos"A"(sin"A" - cos"A")`
= `(1 + sin"A". cos"A")/(sin"A".cos"A")`
= `1/(sin"A".cos"A") + (sin"A".cos"A")/(sin"A".cos"A")`
= `1/sin"A" . 1/cos"A"+ 1`
= sec A.cosec A + 1
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following trigonometric identities.
`cot^2 A cosec^2B - cot^2 B cosec^2 A = cot^2 A - cot^2 B`
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove the following identities:
`cosA/(1 + sinA) + tanA = secA`
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove the following identity :
(secA - cosA)(secA + cosA) = `sin^2A + tan^2A`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
