If 3sec2θ + tanθ = 7, (0° < θ < 90°), then
find the value of \(\frac{cosec2θ\;+\;cosθ}{sin2θ\;+\;cotθ}\)
Answer & explanation
Correct answer: option 3
Put θ = 45°
3(\(\sqrt {2}\))2 + 1 = 7 (satisfied)
⇒ \(\frac{cosec90°\;+\;cos45°}{sin90°\;+\;cot45°}\)
= \(\frac{1+\frac{1}{\sqrt {2}}}{1+1}\)
= \(\frac{\sqrt {2}+1}{2\sqrt {2}}\)
= \(\frac{\sqrt {2}+1}{\sqrt {2}}\)×\(\frac{\sqrt {2}}{\sqrt {2}}\)
= \(\frac{2+\sqrt {2}}{4}\)