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Question
If `sec θ = 41/40`, then find values of sin θ, cot θ, cosec θ.
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Solution
`sec θ = 41/40` ...[Given]
∴ `cos θ = 1/(secθ) = 1/(41/40)`
∴ `cos θ = 40/41`
We know that,
sin2θ + cos2θ = 1
∴ `sin^2θ + (40/41)^2 = 1`
∴ `sin^2θ + 1600/1681 = 1`
∴ `sin^2θ = 1 - 1600/1681`
∴ `sin^2θ = (1681- 1600)/1681`
∴ `sin^2θ = 81/1681`
∴ `sin θ = 9/41` ...[Taking square root of both sides]
Now, cosec θ = `1/(sinθ)`
= `1/((9/41))`
= `41/9`
`cot θ = (cosθ)/(sinθ)`
= `((40/41))/((9/41))`
= `40/9`
∴ `sin θ = 9/41, cot θ = 40/9`, cosec θ = `41/9`
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