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Question
Prove the following.
secθ (1 – sinθ) (secθ + tanθ) = 1
Solution
secθ (1 – sinθ)(secθ + tanθ) = 1
LHS = secθ (1 – sinθ)(secθ + tanθ)
LHS = (secθ – secθ sinθ)(secθ + tanθ)
`"LHS" = (secθ – 1/cosθ × sinθ)(secθ + tanθ)`
`"LHS" = (secθ – sinθ/cosθ)(secθ + tanθ)`
`"LHS" = (secθ – tan θ)(secθ + tanθ) ...[(a + b)(a - b) = a^2 - b^2]`
`"LHS" = sec^2θ – tan^2θ ...{(1 + tan^2θ = sec^2θ),(∴ sec^2θ − tan^2θ = 1):}`
LHS = 1
RHS = 1
LHS = RHS
Hence proved.
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sin \[\theta\] cosec \[\theta\]= ?
Prove the following.
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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
