If 3sec²θ + tanθ = 7, (0° < θ < 90°), then

find the value of \(\frac{cosec2θ\;+\;cosθ}{sin2θ\;+\;cotθ}\)

\(\frac{2+\sqrt {2}}{4}\)

Put θ = 45°

3(\(\sqrt {2}\))² + 1 = 7  (satisfied)

⇒ \(\frac{cosec90°\;+\;cos45°}{sin90°\;+\;cot45°}\)

= \(\frac{1+\frac{1}{\sqrt {2}}}{1+1}\)

= \(\frac{\sqrt {2}+1}{2\sqrt {2}}\)

= \(\frac{\sqrt {2}+1}{\sqrt {2}}\)×\(\frac{\sqrt {2}}{\sqrt {2}}\)

= \(\frac{2+\sqrt {2}}{4}\)