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Question
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
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Solution
`(p^2 - 1)/(p^2 + 1)`
= `((secA + tanA)^2 - 1)/((secA + tanA)^2 + 1)`
= `(sec^2A + tan^2A + 2tanA secA - 1)/(sec^2A + tan^2A + 2tanA secA + 1)`
= `(tan^2A + tan^2A + 2tanA secA)/(sec^2A + sec^2A + 2tanA secA)`
= `(2tan^2A + 2tanA secA)/(2sec^2A + 2tanA secA)`
= `(2tanA(tanA + secA))/(2secA(tanA + secA)`
= sin A
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Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
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and sin θ = `1/("cosec" θ)`
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∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
