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

Answer 1

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