If \(\sqrt {17}\) sinθ = 2, find the value of (\(\frac{\sqrt {13}tanθ + \sqrt {17}sinθ}{cosec^2θ}\))

\(\frac{8}{17}\)

\(\sqrt {17}\) sinθ = 2

sinθ = \(\frac{2}{\sqrt {17}}\)  (where 2 → P and \(\sqrt {17}\) → H)

B = \(\sqrt {(\sqrt {17})^2 - (2)^2}\)

B = \(\sqrt {13}\)

Put all the values →

⇒ (\(\frac{\sqrt {13}tanθ + \sqrt {17}sinθ}{cosec^2θ}\))

= \(\frac{\sqrt {13}×\frac{2}{\sqrt {13}} + \sqrt {17}×\frac{2}{\sqrt {17}}}{(\frac{\sqrt {17}}{2})^2}\)

= \(\frac{4}{17}\) × 2 = \(\frac{8}{17}\)