If \(\sqrt {17}\) sinθ = 2, find the value of (\(\frac{\sqrt {13}tanθ + \sqrt {17}sinθ}{cosec^2θ}\))
Answer & explanation
Correct answer: option 3
\(\sqrt {17}\) 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 = \(\frac{8}{17}\)