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Question
If 5 sec θ – 12 cosec θ = 0, then find values of sin θ, sec θ.
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Solution
5 sec θ – 12 cosec θ = 0 ...[Given]
∴ 5 sec θ = 12 cosec θ
∴ `5/(cosθ) = 12/(sinθ)` ...`[∵ sec θ = 1/(cosθ), "cosec" θ = 1/(sin θ)]`
∴ `(sinθ)/(cosθ) = 12/5`
∴ `tan θ = 12/5`
We know that,
1 + tan2θ = sec2θ
∴ `1 + (12/5)^2 = sec^2θ`
∴ `1 + 144/25 = sec^2θ`
∴ `(25 + 144)/25 = sec^2θ`
∴ `sec^2θ = 169/25`
∴ `sec θ = 13/5` ...[Taking square root of both sides]
Now, `cos θ = 1/(sec θ)`
= `1/((13/5))`
∴ `cos θ = 5/13`
We know that,
sin2θ + cos2θ = 1
∴ `sin^2θ + (5/13)^2 = 1`
∴ `sin^2θ + 25/169 = 1`
∴ `sec^2θ = 1 - 25/169`
∴ `sec^2θ = (169 - 25)/169`
∴ `sec^2θ = 144/169`
∴ `sin θ = 12/13` ...[Taking square root of both sides]
∴ `sin θ = 12/13, sec θ = 13/5`
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