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Question
Prove that: `1/"sec θ − tan θ" = "sec θ + tan θ"`
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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Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`
= `((1 - square)/square) ((square + square)/(square square))`
= `square/square xx 1/(square square)` ......`[(∵ square + square = 1),(∴ square = 1 - square)]`
= `square/(square square)`
= tan θ.sec θ
= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
