(a) \(\sqrt {\frac {1-cosθ }{2}} = \sqrt {\frac {1-(1-2sin^2\frac θ 2)}{2}} = \sqrt {\frac {2 sin^2\frac θ 2}{2}} = sin \frac θ 2\) - it is True.

(b) \(\frac {1+cosθ }{sinθ } = \frac {1+2cos^2\frac θ 2-1}{2\, sin \frac θ 2 cos \,\frac θ 2}= \frac {2\, cos^2\frac θ 2}{2\,sin \frac θ 2 cos\frac θ 2} = cot\frac θ 2 \neq tan \frac θ 2\) -  It is not true.

(c) \(\sqrt {\frac {1-cos \frac {5\pi}6}{2}} = \sqrt {\frac {1-(1-2\, sin^2 \frac {5\pi}{12})}{2}} = sin \frac {5\pi}{12} \neq cos\frac {5\pi}{12}\) - It is false statement.

(d) \(\sqrt {\frac {1-cos 240°}{2}} = \sqrt {\frac {1-(1-2\, sin^2(\frac {240°}{2})}{2}} = sin\, 120°\) - it is true statement.

(e) cot θ = \(\frac {-12}{5}\)

∴ cos θ = \(\frac {12}{\sqrt{(-12)^2 + 5^2}} = \frac {12}{\sqrt {144 + 25}} = \frac {12}{\sqrt{169}} = \frac {12}{13} (∵ \frac {3\pi}{2}< \theta < 2\pi \) lies in 4th quadrant)

∴ 1 - sin² \(\frac θ2\) = \(\frac {12}{13}\) (∵ cos θ =  1-2 sin² \(\frac θ2\) )

= \(2\, sin^2 \frac θ2 = 1- \frac {12}{13}= \frac {1}{13}\) 

= \(\, sin^2 \frac θ2 = \frac {1}{26}\) 

∴ sin \(\frac θ2\) =  \(\frac {1}{\sqrt{26}} = cosec\frac θ2 = \frac {1}{sin\frac θ2} = \sqrt {26}\) - It is true statement.

(f) cos θ = \(\frac {-15}{17} \)        \(\pi < θ < \frac {3\pi}{4}\)

∴ \(\frac {\pi}{2} < \frac θ2 < \frac {3\pi}{8}\) i.e. \( \frac θ2 \) lies in 2nd quadrant.

Now, 1-2 sin²\( \frac θ2 \) = \(\frac {-15}{17}\) (∵ cosθ = 1 - 2 sin²\( \frac θ2 \) )

=  2 sin²\( \frac θ2 \) = 1 + \(\frac {-15}{17}\) = \(\frac {32}{17}\)

=  sin²\( \frac θ2 \) = \(\frac {16}{17}\) 

∴ sin \( \frac θ2 \) = \(\frac {4}{\sqrt{17}}\) (∵ sinθ is positive in 2nd quadrant)

∴ cos \( \frac θ2 \) =  \(\frac {\sqrt{17-16}}{\sqrt{17}} = \frac {-1}{\sqrt 17}\) (∵ cosθ is negative in 2nd quadrant)

∴ tan \( \frac θ2 \) = \(\frac {sin\fracθ 2 }{cos \frac θ 2} = \frac {\frac {4}{\sqrt{17}}}{\frac {-1}{\sqrt{17}}} = -4\) - it is true statement.

(g) sin θ = \(\frac {5}{13}, 0<\theta < 90°\) 

= 0° < \( \frac θ2 \) < 45°

= All trigonometric ratios are positive for \( \frac θ2 \).

Now, cos θ = \(\sqrt{\frac {13^2-5^2}{13}} = \frac {12}{13}\) 

∴ 1-2 sin² \( \frac θ2 \) = \( \frac {12}{13}\) 

=  2 sin² \( \frac θ2 \) = \(1- \frac {12}{13} = \frac 1{13}\) 

= sin \( \frac θ2 \) = \(\sqrt {\frac {1}{26}} = \frac {1}{\sqrt {26}} = \frac {\sqrt{26}}{\sqrt {26}.\sqrt{26}} = \frac {\sqrt{26}}{26}\) -  It is true statement.

(h) tan θ = \(\frac {-3}{4} \)     \(\frac {\pi}{2}<θ <\pi\)

= \(\frac \pi4 < \frac θ 2 < \frac \pi2\)

= \( \frac θ2 \) lies in first quadrant.

Now, \(\frac {2\, tan \frac θ 2​}{\sqrt {1- tan^2 \frac θ 2}} = \frac {-3}{4}\)  (∵ tan θ = \(\frac {2\, tan θ​}{\sqrt {1- tan^2θ}} \) )

= \(\frac {4\, tan^2 \frac θ 2 ​}{1- tan^2 \frac θ 2} = \frac {9}{16} \) (By squaring both sides)

=  64 tan² \( \frac θ2 \) = 9-9  tan² \( \frac θ2 \) 

= 73  tan² \( \frac θ2 \) = 9

=  tan² \( \frac θ2 \) = \(\frac {9}{73}\)

=  tan \( \frac θ2 \) = \(\frac {3}{\sqrt73} \neq -2\)  - Hence, it is false statement.

(i) sin θ = \(\frac {-1}{3} \)       \(\pi < θ < \frac {3\pi}{2}\)

∴ cos θ = \(\frac {\sqrt{3^2-1^2}}{3} = \frac {-2\sqrt2}{3}\) (∵ cosθ is negative in 3rd quadrant)

Also \(\frac {\pi}{2}< \frac θ 2 = \frac {3\pi}{4}\) 

= \( \frac θ2 \) lies in 2nd quadrant

∴ 1-2 sin² \( \frac θ2 \) = \( \frac {-2\sqrt2}{3}\) 

=  2 sin² \( \frac θ2 \) = \( \frac {1+2\sqrt2}{3}\) = \( \frac {3+2\sqrt2}{3}\) 

= sin \( \frac θ2 \) = \(\sqrt { \frac {3+2\sqrt2}{6}}\) 

& 2 cos²\( \frac θ2 \) -1 = \( \frac {-2\sqrt2}{3}\) 

= 2 cos² \( \frac θ2 \) = 1- \( \frac {-2\sqrt2}{3}\) = \( \frac {3-2\sqrt2}{3}\) 

=  cos \( \frac θ2 \) = - \(\sqrt\frac {3-2\sqrt2}{6}\) 

∴  tan \( \frac θ2 \) = \(\frac {sin\frac θ2}{cos \frac θ2} = \frac {\sqrt{\frac{3+2\sqrt2}{\sqrt6}}}{\frac {-\sqrt {3-2\sqrt2}}{\sqrt6}}\) = \(\frac {\sqrt{3+2\sqrt2}}{\sqrt {3-2\sqrt2}} = \frac {-\sqrt{(3+2\sqrt2)^2}}{\sqrt {9-8}} = 3-2\sqrt2\) - It is true statement