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Question
Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
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
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Solution
Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)
= `(1/bbcos θ) - cos θ) (bb cos θ/bb sin θ + bbsin θ/bb cos θ)` ......`[∵ sec θ = 1/bb cos θ, cot θ = bb cos θ/bb sin θ and tan θ = bb sin θ/bb cos θ]`
= `((1 - bb(cos^2 θ))/bbcos θ) ((bb(cos^2 θ) + bb(sin^2 θ))/(bb(sin^2θ) bbcos θ))`
= `bb(sin^2θ)/bb cos θ xx 1/(bbsin θ bbcos θ)` ......`[(∵ bb(sin^2θ) + bb(cos^2θ) = 1),(∴ bb(sin^2θ) = 1 - bb(cos^2θ))]`
= `bbsin θ/(bbcos θ bbcosθ)`
= 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
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