Question

Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`

Answer 1

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

Answer 2

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.