If θ is a positive acute angle and tan2θ tan3θ = 1 find the value of (2cos2\(\frac{5θ}{2}-1\)).

\(\frac{-1}{2}\) 1 0 \(\frac{1}{2}\)

0

If tan2θ tan3θ = 1 than 2θ + 3θ = 90° θ = 18° Put in find out (2cos2\(\frac{18\;×\;5}{2}\;-\;1\)) = 2 × (\(\frac{1}{\sqrt {2}}\))2 - 1 = 2 × \(\frac{1}{2}\) - 2 = 0