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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)`
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Solution 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.
Solution 2
LHS = `(1-sectheta + tantheta)/(1+sectheta-tantheta)`
Denominator = (a − b) (a + b) = a2 − b2
Let a = 1 + secθ, b = tanθ
Denominator = (1 + secθ)2 − tan2θ
= 1 + 2secθ + sec2θ − tan2θ
= 1 + 2secθ + 1 = 2(1 + secθ)
denominator = 2(1 + secθ)
Numerator = (1 − secθ + tanθ) (1 + secθ + tanθ)
= 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 sec2θ − tan2θ = 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)`
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