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Question
If 5 secθ – 12 cosecθ = 0, find the values of secθ, cosθ, and sinθ.
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Solution
5secθ - 12cosecθ = 0
⇒ 5secθ = 12cosecθ
⇒ `5xx1/cosθ=12xx1/sinθ`
⇒ `5/cosθ = 12/sinθ`
⇒ `sinθ/cosθ = 12/5`
⇒ tanθ = `12/5 ...[tanθ=sinθ/cosθ]`
We have,
sec2θ = 1 + tan2θ
⇒ sec2θ = 1 + `(12/5)^2`
⇒ sec2θ = 1 + `144/25`
⇒ sec2θ = `169/25`
Taking square root on both sides,
`sqrt(sec^2θ)=sqrt(169/25)`
∴ secθ = `13/5`
Now,
cosθ = `1/secθ`
⇒ cosθ = `1/(13/5)`
⇒ cosθ = `5/13`
Also,
`sinθ/cosθ` = tanθ
⇒ sinθ = tanθ × cosθ
⇒ sinθ = `12/5 xx 5/13 = 12/13`
Thus, the values of secθ, cosθ and sinθ are `13/5, 5/13 and 12/13` respectively.
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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))`
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= `square/(square square)`
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= R.H.S.
∴ L.H.S. = R.H.S.
∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ
