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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)`
Solution
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.
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