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If sec θ + tan θ = p, then tan θ is(a) \(\frac{p^2 + 1}{2p}\)(b) \(\frac{p^2 - 1}{2p}\)(c) \(\frac{p^2 - 1}{p^2 + 1}\)(d) \(\frac{p^2 + 1}{p^2 - 1}\) |
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Answer» Correct answer is (b) \(\frac{p^2 - 1}{2p}\) |
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