Question

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QUESTION OF THE DAY
If tanθ+secθ=√3,0<θ<π, then θ is equal to
π
3
π
6
2π
3
5π
6

Answer 1

given that

$tan \theta + sec \theta = \sqrt{3} \: \: \: \: \: \: \: \: \: ... \: (i) \\ \\</p><p>we \: know \: that; \\ \\</p><p>1 + tan{}^{2} \theta = sec{}^{2} \theta \\</p><p>sec{}^{2} \theta - tan{}^{2} \theta = 1 \\</p><p>(sec \theta - tan \theta)(sec \theta + tan \theta) = 1 \\</p><p>(sec \theta - tan \theta)(\sqrt{3}) = 1 \: \: \: \: \: \: \: \: \: ... \: (from i) \\</p><p>sec \theta - tan \theta = \frac{1}{\sqrt{3}} \: \: \: \: \: \: \: \: \: ... \: (ii) \\ \\</p><p>adding \: (i) \: and \: (ii), \: we \: get; \\ \\</p><p>2sec \theta = \sqrt{3} + \frac{1}{\sqrt{3}} \\</p><p>2sec \theta = \frac{3 + 1}{\sqrt{3}} \\</p><p>sec \theta = \frac{4}{2\sqrt{3}} \\</p><p>sec \theta = \frac{2}{\sqrt{3}} \\</p><p>sec \theta = sec(\frac{ \pi}{6} )\\</p><p>\theta = \frac{ \pi}{6}$