Advertisements
Advertisements
प्रश्न
Prove that `sqrt((1 - sin θ)/(1 + sin θ)) = sec θ - tan θ`.
Advertisements
उत्तर १
L.H.S. = `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ)))`
= `sqrt((1 + sin^2θ - 2sinθ)/(1 - sin^2θ)`
= `sqrt((1 + sin^2θ - 2sinθ)/(cos^2θ)`
= `sqrt( 1/cos^2θ + sin^2θ/cos^2θ - (2sin θ)/cos θ xx 1/cosθ`
= `sqrt( sec^2θ + tan^2 θ - 2 tan θ . sec θ)`
= `sqrt((sec θ - tan θ)^2)`
= sec θ – tan θ
= R.H.S.
Hence proved.
उत्तर २
L.H.S. = `sqrt((1 - sin θ)/(1 + sin θ))`
= `sqrt(((1 - sin θ)(1 - sin θ))/((1 + sin θ)(1 - sin θ))`
= `sqrt(((1 - sin θ)^2)/(1 - sin^2θ)`
= `sqrt(((1 - sin θ)^2)/(cos^2θ)`
= `(1 - sin θ)/(cos θ)`
= `1/(cos θ) - (sin θ)/(cos θ)`
= sec θ – tan θ
= R.H.S.
Hence Proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
Prove that:
`sqrt(sec^2A + cosec^2A) = tanA + cotA`
` tan^2 theta - 1/( cos^2 theta )=-1`
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Prove the following identities:
`(sec"A"-1)/(sec"A"+1)=(sin"A"/(1+cos"A"))^2`
Prove that sin4θ - cos4θ = sin2θ - cos2θ
= 2sin2θ - 1
= 1 - 2 cos2θ
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt(a^2 + b^2 - c^2)`
cos θ . sec θ = ?
Prove that sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A.
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
