If $cosec θ =\frac{\sqrt{5}}{2}$, then what will be the value of $(sec θ + tan θ- cotθ sinθ)$ ?

$2 + \frac{4\sqrt{5}}{5}$

$cosec θ =\frac{\sqrt{5}}{2}$,

cosecθ = \(\frac{H}{P}\)

By using pythagoras theorem,

P² + B² = H²

2² + B² = √5²

B = 1

Now,

sec θ+ tanθ - cotθ. sinθ

= \(\frac{√5}{1}\) + \(\frac{2}{ 1 }\) - \(\frac{1}{2}\) . \(\frac{2}{√5}\)

= \(\frac{2 +√5 }{1}\) - \(\frac{1}{√5}\)

= \(\frac{2 +√5 }{1}\) - \(\frac{1}{√5}\)

= \(\frac{2√5 +5 - 1 }{√5}\)

= 2 + \(\frac{4√5 }{5}\)