Answer:
Step-by-step explanation:
tan \theta=12/5
[tex]}}{6}sec^2 \theta-tan ^2 \theta=1\\sec^2 \theta=1+tan ^2 \theta\\=1+(\frac{12}{5})^2\\=1+\frac {144}{25} \\=\frac {169}{25}\\cos^2 \theta=\frac {25}{169}\\cos \theta=-\frac {5}{13} ~as~\pi <\theta ~\leq \frac{3\pi }{2}\\1+cos \theta=1-\frac{5}{13}=\frac{8}{13}\\2 cos^2 (\frac{\theta}{2} )=\frac{8}{13} \\cos \frac{\theta}{2}=-\frac{2}{\sqrt{13}}\\2sin ^2 \theta=1-cos 2 \theta\\put ~\theta=\frac{5\pi }{12}\\2 \theta=\frac {5\pi }{6}\\1-cos~2 \theta=1-cos\frac{5\pi }{6} =1-cos(\pi -\frac{\pi }{6} )\\[/tex]
=1+cos (π/6)
=1+(√3)/2
=(2+√3)/2
2sin^2(5π/12)=(2+√3)/2
sin (5π/12)=(√(2+√3))/2
