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Question
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
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Solution
LHS = `1/"sec θ − tan θ"`
LHS = `1/"sec θ − tan θ" × "sec θ + tan θ"/"sec θ + tan θ"`
LHS = `"sec θ + tan θ"/((sec θ − tan θ)(sec θ + tan θ))`
LHS = `(sec θ + tan θ)/(sec^2θ − tan^2θ) ...[(a + b)(a - b) = a^2 - b^2]`
LHS = `(sec θ + tan θ)/1 ...{(1 + tan^2θ = sec^2θ),(∴ sec^2θ − tan^2θ = 1):}`
LHS = sec θ + tan θ
RHS = sec θ + tan θ
LHS = RHS
Hence proved.
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