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If cosec θ $=\frac{41}{9}$ and θ is an acute angle, then the value of 5 tan θ will be: |
$\frac{11}{8}$ $\frac{13}{4}$ $\frac{7}{8}$ $\frac{9}{8}$ |
$\frac{9}{8}$ |
cosec θ = \(\frac{41 }{9}\) = \(\frac{H }{P}\) B y using pythogoras theorem , H² = B² + P² (41)² = B² + (9)² B² = 1681 - 81 = 1600 B = 40 Now , 5 tan θ = 5 × \(\frac{9 }{40}\) ( tan θ = \(\frac{P }{B}\) ) = \(\frac{9 }{8}\)
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