Advertisements
Advertisements
प्रश्न
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Advertisements
उत्तर १
L.H.S = `(tan^2 theta)/(sec theta - 1)^2 `
`= (sec^2 theta - 1)/(sec theta - 1)^2`
`= ((sec theta - 1)(sec theta + 1))/(sec theta - 1)^2`
`= (sec theta + 1)/(sec theta - 1)`
`= (1/cos theta + 1)/(1/cos theta - 1)`
`= ((1 + cos theta)/cos theta)/((1 - cos theta)/cos theta)`
`= (1 + cos theta)/(1 - cos theta)`
= R.H.S
उत्तर २
L.H.S = `(tan^2 θ)/(sec θ - 1)^2 `
= `(sin^2 θ/cos^2 θ)/(1/cos θ - 1)^2 .....( ∵ tan θ = sin θ /cos θ )`
= `(sin^2 θ/cos^2 θ)/((1/cos θ - 1)^2/(cos^2 θ)) ....( ∵ sec θ = 1/cos θ) `
= `(sin^2 θ)/( 1 - cos θ)^2 ....( ∵ sin^2 θ = 1 - cos^2 θ) `
= `( 1 - cos^2 θ)/( 1 - cos θ)^2`
= `( 1 - cos θ)( 1 + cos θ)/( 1 - cos θ)^2` ....( ∵ a2 - b2 = (a + b)(a - b))
= `( 1 + cos θ)/( 1 - cos θ)`
= RHS.
APPEARS IN
संबंधित प्रश्न
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:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
Prove the following identity :
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
If sin θ = `1/2`, then find the value of θ.
cos 45° = ?
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = `sqrt(a^2 + b^2 - c^2)`.
