Question

Prove the following identity:

`(1 - sec theta + tan theta)/(1 + sec theta - tan theta) = (sec theta + tan theta - 1)/(sec theta + tan theta + 1)`

Answer 1

1 + tan² θ = sec² θ

∴ sec² θ − tan² θ = 1

Let, a² − b² = (a + b)(a - b) so,

∴ (sec θ + tan θ) (sec θ − tan θ) = 1

∴ `(sec θ + tan θ) = 1/(sec θ + tan θ)`

∴ `(sec θ + tan θ)/1 = 1/(sec θ + tan θ)`

By componendo-dividendo, we get

∴ `(sec theta + tan theta + 1)/(sec theta + tan theta - 1) = (1 + sec theta - tan theta)/(1 - (sec theta - tan theta))`

∴ `(sec theta + tan theta + 1)/(sec theta + tan theta - 1) = (1 + sec theta - tan theta)/(1 - sec theta + tan theta)`

∴ `(1 + sec theta - tan theta)/(1 - sec theta + tan theta) = (sec theta + tan theta + 1)/(sec theta + tan theta - 1)`

By Invertendo, we get

∴ `(1 - sec theta + tan theta)/(1 + sec theta - tan theta) = (sec theta + tan theta - 1)/(sec theta + tan theta + 1)`

Hence proved. 

Answer 2

LHS = `(1-sectheta + tantheta)/(1+sectheta-tantheta)`

Denominator = (a − b) (a + b) = a² − b²

Let a = 1 + sec⁡θ, b = tan⁡θ

Denominator = (1 + secθ)² − tan²θ

= 1 + 2secθ + sec²θ − tan²θ

= 1 + 2secθ + 1 = 2(1 + secθ)

denominator = 2(1 + sec⁡θ)

Numerator = (1 − sec⁡θ + tan⁡θ) (1 + sec⁡θ + tan⁡θ)

= 1(1 + secθ + tanθ) − secθ (1 + secθ + tanθ) + tanθ (1 + secθ + tanθ)

= (1 + secθ + tanθ) − (secθ + sec²θ + secθ tanθ) + (tanθ + secθ tanθ + tan²θ)

= 1 − sec²θ + tan²θ + 2tanθ

Use the identity sec⁡²θ − tan⁡²θ = 1

1 − (sec²θ − tan²θ) + 2tanθ = 1 − 1 + 2tanθ = 2tanθ

`LHS = (2tantheta)/(2(1+sectheta)) = tantheta/(1+sectheta)`

RHS = `(sectheta+tantheta-1)/(sectheta + tantheta+1)`

`= (1-(1/(sectheta+tantheta)))/(1+(1/(sectheta+tantheta))`

Use the standard result: `(1-x)/(1+x) = (1-x)/(1+x)`

`tantheta/(1+sectheta) = (sectheta+tantheta-1)/(sectheta+tantheta+1)`

Hence proved `(1 - sec theta + tan theta)/(1 + sec theta - tan theta) = (sec theta + tan theta - 1)/(sec theta + tan theta + 1)`