Question 1: Find the values of the following trigonometric ratios:

$\displaystyle \text{(i) } \sin \Big( \frac{5 \pi}{3} \Big) \hspace{1.0cm} \text{(ii) } \sin 17 \pi \hspace{1.0cm} \text{(iii) } \tan \Big( \frac{11 \pi}{6} \Big) \hspace{1.0cm} \\ \\ \text{(iv) } \cos \Big( - \frac{25\pi}{4} \Big) \hspace{1.0cm} \text{(v) } \tan \Big( \frac{7\pi}{4} \Big) \hspace{1.0cm} \text{(vi) } \sin \Big( \frac{17\pi}{6} \Big) \hspace{1.0cm} \\ \\ \text{(vii) } \cos \Big( \frac{19\pi}{6} \Big) \hspace{1.0cm} \text{(viii) } \sin \Big( - \frac{11\pi}{6} \Big) \hspace{1.0cm} \text{(ix) } \mathrm{cosec} \Big( - \frac{20\pi}{3} \Big) \hspace{1.0cm} \\ \\ \text{(x) } \tan \Big( - \frac{13\pi}{4} \Big) \hspace{1.0cm} \text{(xi) } \cos \Big( \frac{19\pi}{4} \Big) \hspace{1.0cm} \text{(xii) } \sin \Big( \frac{41\pi}{4} \Big) \hspace{1.0cm} \\ \\ \text{(xiii) } \cos \Big( \frac{39\pi}{4} \Big) \hspace{1.0cm} \text{(xiv) } \sin \Big( \frac{151\pi}{6} \Big)$

$\displaystyle \text{(i) } \sin \Big( \frac{5 \pi}{3} \Big) = \sin \Big( 2\pi - \frac{\pi}{3} \Big) = - \sin \frac{\pi}{3} = - \frac{\sqrt{3}}{2}$

$\displaystyle \text{(ii) } \sin 17 \pi = 0 (\because n\pi = 0$ for all $\displaystyle n \in Z)$

$\displaystyle \text{(iii) } \tan \Big( \frac{11 \pi}{6} \Big) = \tan \Big( 2\pi - \frac{\pi}{6} \Big) = - \tan \frac{\pi}{6} = - \frac{1}{\sqrt{3}}$

$\displaystyle \text{(iv) } \cos \Big( - \frac{25\pi}{4} \Big) = \cos \Big( 6\pi + \frac{\pi}{4} \Big) = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ [Note: $\displaystyle \cos(-x) = \cos x$ ]

$\displaystyle \text{(v) } \tan \Big( \frac{7\pi}{4} \Big) = \tan \Big( 2\pi - \frac{\pi}{4} \Big) = - \tan \frac{\pi}{4} = -1$

$\displaystyle \text{(vi) } \sin \Big( \frac{17\pi}{6} \Big) = \sin \Big( 3\pi - \frac{\pi}{6} \Big) = \sin \frac{\pi}{6} = \frac{1}{2}$

$\displaystyle \text{(vii) } \cos \Big( \frac{19\pi}{6} \Big) = \cos \Big( 3\pi + \frac{\pi}{6} \Big) = -\cos \frac{\pi}{6} = - \frac{\sqrt{3}}{2}$

$\displaystyle \text{(viii) } \sin \Big( - \frac{11\pi}{6} \Big) = \sin \Big(- \Big( 2\pi - \frac{\pi}{6} \Big) \Big) = \sin \Big( 2\pi - \frac{\pi}{6} \Big) = \sin \frac{\pi}{6} = \frac{1}{2}$

$\displaystyle \text{(ix) } \mathrm{cosec} \Big( - \frac{20\pi}{3} \Big) = \mathrm{cosec} \Big(- \Big( 7\pi - \frac{\pi}{3} \Big) \Big) = - \mathrm{cosec} \Big( 7\pi - \frac{\pi}{3} \Big) = -\mathrm{cosec} \frac{\pi}{3} = - \frac{2}{\sqrt{3}}$

$\displaystyle \text{(x) } \tan \Big( - \frac{13\pi}{4} \Big) = - \tan \Big( 3\pi + \frac{\pi}{4} \Big) = - \tan \Big( 2\pi +\pi + \frac{\pi}{4} \Big) = - \tan \frac{\pi}{4} = -1$

$\displaystyle \text{(xi) } \cos \Big( \frac{19\pi}{4} \Big) = \cos \Big( 5\pi - \frac{\pi}{4} \Big) = \cos \Big( 4\pi +\pi - \frac{\pi}{4} \Big) = \cos \Big( \pi - \frac{\pi}{4} \Big) = - \cos \frac{\pi}{4} = - \frac{1}{2}$

$\displaystyle \text{(xii) } \sin \Big( \frac{41\pi}{4} \Big) = \sin \Big( 10\pi + \frac{\pi}{4} \Big) = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\displaystyle \text{(xiii) } \cos \Big( \frac{39\pi}{4} \Big) = \cos \Big( 10\pi - \frac{\pi}{4} \Big) = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\displaystyle \text{(xiv) } \sin \Big( \frac{151\pi}{6} \Big) = \sin \Big( 25\pi + \frac{\pi}{6} \Big) = \sin \Big( 24\pi +\pi + \frac{\pi}{6} \Big) = \sin \Big( \pi + \frac{\pi}{6} \Big) = - \sin \frac{\pi}{6} = - \frac{1}{2}$

$\displaystyle \\$

Question 2: Prove that:

$\displaystyle \text{(i) } \tan 225^{\circ} \cot 405^{\circ} + \tan 765^{\circ} \cot 675^{\circ} = 0$

$\displaystyle \text{(ii) } \sin \frac{8 \pi}{3} \cos \frac{23 \pi}{6} + \cos \frac{13 \pi}{3} \sin \frac{35 \pi}{6} = \frac{1}{2}$

$\displaystyle \text{(iii) } \cos 24^{\circ} + \cos 55^{\circ} + \cos 125^{\circ} + \cos 204^{\circ} + \cos 300^{\circ} = \frac{1}{2}$

$\displaystyle \text{(iv) } \tan (-225^{\circ}) \cot(-405^{\circ}) - \tan (-765^{\circ}) \cot (675^{\circ}) = 0$

$\displaystyle \text{(v) } \cos 570^{\circ} \sin 510^{\circ} + \sin (-300^{\circ}) \cos (-390^{\circ}) = 0$

$\displaystyle \text{(vi) } \tan \frac{11 \pi}{3} - 2 \sin \frac{4 \pi}{6} - \frac{3}{4} \mathrm{cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3-4\sqrt{3}}{2}$

$\displaystyle \text{(vii) } 3\sin \frac{\pi}{6} \sec \frac{ \pi}{3} -4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4} = 1$

$\displaystyle \text{(i) } \text{LHS } = \tan 225^{\circ} \cot 405^{\circ} + \tan 765^{\circ} \cot 675^{\circ}$

$\displaystyle = \tan \Big( \pi + \frac{\pi}{4} \Big) \cot \Big( 2\pi + \frac{\pi}{4} \Big) + \tan \Big( 4\pi + \frac{\pi}{4} \Big) \cot \Big( 4\pi - \frac{\pi}{4} \Big)$

$\displaystyle = \tan \frac{\pi}{4} \cot \frac{\pi}{4} + \tan \frac{\pi}{4} \Big( - \cot \frac{\pi}{4} \Big)$

$\displaystyle = 1 - 1 = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{(ii) } \text{LHS } = \sin \frac{8 \pi}{3} \cos \frac{23 \pi}{6} + \cos \frac{13 \pi}{3} \sin \frac{35 \pi}{6}$

$\displaystyle = \sin \Big( 3\pi - \frac{\pi}{3} \Big) \cos \Big( 4\pi - \frac{\pi}{6} \Big) + \cos \Big( 4\pi + \frac{\pi}{3} \Big) \sin \Big( 6\pi - \frac{\pi}{6} \Big)$

$\displaystyle = \sin \frac{\pi}{3} \cos \frac{\pi}{6} + \cos \frac{\pi}{3} \Big( - \sin \frac{\pi}{6} \Big)$

$\displaystyle = \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \Big( - \frac{1}{2} \Big)$

$\displaystyle = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$

$\displaystyle \text{(iii) } \text{LHS } = \cos 24^{\circ} + \cos 55^{\circ} + \cos 125^{\circ} + \cos 204^{\circ} + \cos 300^{\circ}$

$\displaystyle = ( \cos 24^{\circ} + \cos 204^{\circ}) + ( \cos 55^{\circ} + \cos 125^{\circ}) + \cos 300^{\circ}$

$\displaystyle = ( \cos 24^{\circ} + \cos (\pi + 24^{\circ}) ) + ( \cos 55^{\circ} + \cos (\pi - 55^{\circ})) + \cos (2\pi - 60^{\circ})$

$\displaystyle = \cos 24^{\circ} - \cos 24^{\circ} + \cos 55^{\circ} - \cos 55^{\circ} + \cos 60^{\circ}$

$\displaystyle = \cos 60^{\circ} = \frac{1}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \text{(iv) } \text{LHS } = \tan (-225^{\circ}) \cot(-405^{\circ}) - \tan (-765^{\circ}) \cot (675^{\circ})$

$\displaystyle = -\tan 225^{\circ} (-\cot 405^{\circ} ) + \tan 765^{\circ} \cot 675^{\circ}$

$\displaystyle = \tan \Big( \pi + \frac{\pi}{4} \Big) \cot \Big( 2\pi + \frac{\pi}{4} \Big) + \tan \Big( 4\pi + \frac{\pi}{4} \Big) \cot \Big( 4\pi - \frac{\pi}{4} \Big)$

$\displaystyle = \tan \frac{\pi}{4} \cot \frac{\pi}{4} + \tan \frac{\pi}{4} \Big( - \cot \frac{\pi}{4} \Big)$

$\displaystyle = 1 - 1 = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{(v) } \text{LHS } = \cos 570^{\circ} \sin 510^{\circ} + \sin (-300^{\circ}) \cos (-390^{\circ})$

$\displaystyle = \cos \Big( 3\pi + \frac{\pi}{6} \Big) \sin \Big( 3\pi - \frac{\pi}{6} \Big) - \sin \Big( 2\pi - \frac{\pi}{6} \Big) \cos \Big( 2\pi + \frac{\pi}{6} \Big)$

$\displaystyle = - \cos \frac{\pi}{6} \sin \frac{\pi}{6} + \sin \frac{\pi}{6} \cos \frac{\pi}{6}$

$\displaystyle = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{(vi) } \text{LHS } = \tan \frac{11 \pi}{3} - 2 \sin \frac{4 \pi}{6} - \frac{3}{4} \mathrm{cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6}$

$\displaystyle = \tan \Big( 4\pi - \frac{\pi}{3} \Big) - 2 \sin \frac{2\pi}{3} - \frac{3}{4} \mathrm{cosec}^2 \frac{\pi}{4} + 4 \cos^2 \Big( 3\pi - \frac{\pi}{6} \Big)$

$\displaystyle = - \tan \frac{\pi}{3} - 2 \sin \Big( \pi - \frac{\pi}{3} \Big) - \frac{3}{4} (\sqrt{2})^2 + 4 \cos^2 \frac{\pi}{3}$

$\displaystyle = - \sqrt{3} - 2 \frac{\sqrt{3}}{2} - \frac{3}{2} + 4. \frac{3}{4}$

$\displaystyle = - \sqrt{3} - \sqrt{3} + \frac{3}{2} = \frac{3-4\sqrt{3}}{2} =$ RHS.

$\displaystyle \text{(vi) } \text{LHS } = 3\sin \frac{\pi}{6} \sec \frac{ \pi}{3} -4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}$

$\displaystyle = 3 \times \frac{1}{2} \times 2 - 3 \times \sin \Big( \pi - \frac{\pi}{6} \Big) . 1$

$\displaystyle = 3 - 4 \sin \frac{\pi}{6}$

$\displaystyle = 3 - 4 \times \frac{1}{2} = 3 - 2 = 1 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 3: Prove that:

$\displaystyle \text{(i) } \frac{\cos (2 \pi +x) \mathrm{cosec} (2 \pi +x) \tan (\frac{\pi}{2}+x)}{\sec (\frac{\pi}{2}+x) \cos x \cot (\pi + x) } = 1$

$\displaystyle \text{(ii) } \frac{ \mathrm{cosec} (90^{\circ}+x) + \cot(450^{\circ} + x)}{ \mathrm{cosec} (90^{\circ}-x) + \tan (180^{\circ} - x)} + \frac{\tan (180^{\circ}+x)+ \sec (180^{\circ}-x)}{\tan (360^{\circ}+x)- \sec(-x)} = 2$

$\displaystyle \text{(iii) } \frac{\sin (\pi + x) \cos (\frac{\pi}{2} +x ) \tan (\frac{3\pi}{2} -x ) \cot (2\pi-x) }{\sin (2\pi -x) \cos (2\pi +x) \mathrm{cosec} (-x) \sin (\frac{3\pi}{2} -x ) } = 1$

$\displaystyle \text{(iv) } \Big\{ 1 + \cot x - \sec \Big( \frac{\pi}{2} +x \Big) \Big\} \Big\{ 1 + \cot x + \sec \Big( \frac{\pi}{2} +x \Big) \Big\} = 2 \cot x$

$\displaystyle \text{(v) } \frac{\tan ( \frac{\pi}{2}-x ) \sec (\pi - x) \sin (-x) }{ \sin (\pi + x) \cot (2 \pi - x) \mathrm{cosec} ( \frac{\pi}{2}-x )} = 1$

$\displaystyle \text{(i) } \text{LHS } = \frac{\cos (2 \pi +x) \mathrm{cosec} (2 \pi +x) \tan (\frac{\pi}{2}+x)}{\sec (\frac{\pi}{2}+x) \cos x \cot (\pi + x) }$

$\displaystyle = \frac{\cos x \ \mathrm{cosec} x \ (-\cot x)}{- \mathrm{cosec} x \ \cos x \ \cot x }$

$\displaystyle = 1 =$ RHS

$\displaystyle \text{(ii) } \text{LHS } = \frac{ \mathrm{cosec} (90^{\circ}+x) + \cot(450^{\circ} + x)}{ \mathrm{cosec} (90^{\circ}-x) + \tan (180^{\circ} - x)} + \frac{\tan (180^{\circ}+x)+ \sec (180^{\circ}-x)}{\tan (360^{\circ}+x)- \sec(-x)}$

$\displaystyle = \frac{ \sec x + \cot(2\pi + \frac{\pi}{2} + x)}{ \sec x - \tan x} + \frac{\tan x - \sec x}{\tan x- \sec x}$

$\displaystyle = \frac{ \sec x + \cot( \frac{\pi}{2} + x)}{ \sec x - \tan x} + 1$

$\displaystyle = \frac{ \sec x - \tan x}{ \sec x - \tan x} + 1$

$\displaystyle = 1 + 1 = 2 = \text{ RHS. Hence proved. }$

$\displaystyle \text{(iii) } \text{LHS } = \frac{\sin (\pi + x) \cos (\frac{\pi}{2} +x ) \tan (\frac{3\pi}{2} -x ) \cot (2\pi-x) }{\sin (2\pi -x) \cos (2\pi +x) \mathrm{cosec} (-x) \sin (\frac{3\pi}{2} -x ) }$

$\displaystyle = \frac{\sin x (- \sin x ) \cot x (- \cot x ) }{- \sin x \cos x ( - \mathrm{cosec} x ) (- \cos x) }$

$\displaystyle = \frac{\sin^2 x \cot^2 x}{\cos^2 x}$

$\displaystyle = \tan^2 x \cot^2 x$

$\displaystyle = 1= \text{ RHS. Hence proved. }$

$\displaystyle \text{(iv) } \text{LHS } = \Big\{ 1 + \cot x - \sec \Big( \frac{\pi}{2} +x \Big) \Big\} \Big\{ 1 + \cot x + \sec \Big( \frac{\pi}{2} +x \Big) \Big\}$

$\displaystyle = ( 1 + \cot x - (\mathrm{cosec} x) )( 1 + \cot x - \mathrm{cosec} x )$

$\displaystyle = ( 1 + \cot x + \mathrm{cosec} x )( 1 + \cot x - \mathrm{cosec} x )$

$\displaystyle = ( 1 + \cot^2 x )^2 - \mathrm{cosec}^2 x$

$\displaystyle = 1 + \cot^2 x + 2 \cot x - \mathrm{cosec}^2 x$

$\displaystyle = \mathrm{cosec}^2 x + 2 \cot x - \mathrm{cosec}^2 x$

$\displaystyle = 2 \cot x = \text{ RHS. Hence proved. }$

$\displaystyle \text{(v) } \text{LHS } = \frac{\tan ( \frac{\pi}{2}-x ) \sec (\pi - x) \sin (-x) }{ \sin (\pi + x) \cot (2 \pi - x) \mathrm{cosec} ( \frac{\pi}{2}-x )}$

$\displaystyle = \frac{\cot x (- \sec x) (- \sin x) }{ (- \sin x) (- \cot x) \sec x }$

$\displaystyle = 1 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 4: Prove that: } \sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9} = 2$

$\displaystyle \text{LHS } = \sin^2 \frac{\pi}{18} + \sin^2 \frac{\pi}{9} + \sin^2 \frac{7\pi}{18} + \sin^2 \frac{4\pi}{9}$

$\displaystyle = \sin^2 \Big( \frac{\pi}{2} - \frac{4\pi}{9} \Big) + \sin^2 \frac{\pi}{9} + \sin^2 \Big( \frac{\pi}{2} - \frac{\pi}{9} \Big) + \sin^2 \frac{4\pi}{9}$

$\displaystyle = \cos^2 \frac{4\pi}{9} + \sin^2 \frac{4\pi}{9} + \sin^2 \frac{\pi}{9} + \cos^2 \frac{\pi}{9}$

$\displaystyle = 1 + 1 = 2 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 5: Prove that:

$\displaystyle \sec \Big( \frac{3\pi}{2} -x \Big) \sec \Big( x - \frac{5\pi}{2} \Big) + \tan \Big( \frac{5\pi}{2} +x \Big) \tan \Big( x - \frac{3\pi}{2} \Big) = -1$

$\displaystyle \text{LHS } = \sec \Big( \frac{3\pi}{2} -x \Big) \sec \Big( x - \frac{5\pi}{2} \Big) + \tan \Big( \frac{5\pi}{2} +x \Big) \tan \Big( x - \frac{3\pi}{2} \Big)$

$\displaystyle = \sec \Big( \frac{3\pi}{2} -x \Big) \sec \Big( - ( \frac{5\pi}{2} - x \Big) + \tan \Big( \frac{5\pi}{2} +x \Big) \tan \Big( -( \frac{3\pi}{2} - x) \Big)$

$\displaystyle = -\mathrm{cosec} x \sec \Big( \frac{5\pi}{2} - x \Big) - \cot x \Big[ - \tan \Big( \frac{3\pi}{2} - x \Big) \Big]$

$\displaystyle = -\mathrm{cosec} x . \mathrm{cosec} x + \cot x . \cot x$

$\displaystyle = -\mathrm{cosec}^x + \cot^2 x$

$\displaystyle = -\mathrm{cosec}^x + \mathrm{cosec}^x - 1$

$\displaystyle = - 1 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 6: In a $\displaystyle \triangle ABC$, prove that: $\displaystyle \text{(i) } \cos (A+B) + \cos C = 0$

$\displaystyle \text{(ii) } \cos \Big( \frac{A+B}{2} \Big) = \sin \Big( \frac{C}{2} \Big) \hspace{1.0cm} \text{(iii) } \tan \Big( \frac{A+B}{2} \Big) = \cot \Big( \frac{C}{2} \Big)$

$\displaystyle \text{(i) } \text{LHS } = \cos (A+B) + \cos C$

$\displaystyle = \cos (A+B) + \cos (\pi - (A + B))$

$\displaystyle = \cos (A+B) - \cos (A+B)$

$\displaystyle = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \text{(ii) } \text{LHS } = \cos \Big( \frac{A+B}{2} \Big) = \cos \Big( 90^{\circ} - \frac{C}{2} \Big) = \sin \frac{C}{2}$ Hence proved.

$\displaystyle \text{(iii) } \text{LHS } = \tan \Big( \frac{A+B}{2} \Big) = \tan \Big( 90^{\circ} - \frac{C}{2} \Big) = \cot \frac{C}{2}$ Hence proved.

$\displaystyle \\$

Question 7: If A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that: $\displaystyle \cos (180^{\circ} - A) + \cos (180^{\circ} + B) + \cos (180^{\circ} + C ) - \sin (90^{\circ} + D) = 0$

Given $\displaystyle A + C = 180^{\circ}, B+D = 180^{\circ}$

$\displaystyle \text{LHS } = \cos (180^{\circ} - A) + \cos (180^{\circ} + B) + \cos (180^{\circ} + C ) - \sin (90^{\circ} + D)$

$\displaystyle = \cos C + \cos D - \cos C - \cos D = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

Question 8: Find $\displaystyle x$ from the following equations:

$\displaystyle \text{(i) } \mathrm{cosec} \Big( \frac{\pi}{2} + \theta \Big) + x \cos \theta \cot \Big( \frac{\pi}{2} + \theta \Big) = \sin \Big( \frac{\pi}{2} + \theta \Big)$

$\displaystyle \text{(ii) } x \cot \Big( \frac{\pi}{2} + \theta \Big) + \tan \Big( \frac{\pi}{2} + \theta \Big) \sin \theta + \mathrm{cosec} \Big( \frac{\pi}{2} + \theta \Big) = 0$

$\displaystyle \text{(i) } \mathrm{cosec} \Big( \frac{\pi}{2} + \theta \Big) + x \cos \theta \cot \Big( \frac{\pi}{2} + \theta \Big) = \sin \Big( \frac{\pi}{2} + \theta \Big)$

$\displaystyle \Rightarrow \sec \theta + x \cos \theta ( - \tan \theta) = \cos \theta$

$\displaystyle \Rightarrow \frac{1}{\cos \theta} + x ( - \sin \theta) = \cos \theta$

$\displaystyle \Rightarrow \frac{1 - \cos^2 \theta}{\cos \theta} = x \sin \theta$

$\displaystyle \Rightarrow x = \frac{\sin^2 \theta}{\cos \theta . \sin \theta} = \frac{\sin \theta}{\cos \theta} = \tan \theta$

$\displaystyle \text{(ii) } x \cot \Big( \frac{\pi}{2} + \theta \Big) + \tan \Big( \frac{\pi}{2} + \theta \Big) \sin \theta + \mathrm{cosec} \Big( \frac{\pi}{2} + \theta \Big) = 0$

$\displaystyle \Rightarrow x ( - \tan \theta) - \cot \theta \sin \theta + \sec \theta = 0$

$\displaystyle \Rightarrow x \tan \theta = \frac{1}{\cos \theta} - \cos \theta$

$\displaystyle \Rightarrow x \tan \theta = \frac{1 - \cos^2 \theta}{\cos \theta}$

$\displaystyle \Rightarrow x = \frac{\sin^2 \theta . \cos \theta}{\sin \theta . \cos \theta} = \sin \theta$

$\displaystyle \\$

Question 9: Prove that:

$\displaystyle \text{(i) } \tan 4\pi - \cos \frac{3\pi}{2} - \sin \frac{5\pi}{6} \cos \frac{2\pi}{3} = \frac{1}{4}$

$\displaystyle \text{(ii) } \sin \frac{13\pi}{3} \sin \frac{8\pi}{3} + \cos \frac{2\pi}{3} \sin \frac{5\pi}{6} = \frac{1}{2}$

$\displaystyle \text{(iii) } \sin \frac{13\pi}{3} \sin \frac{2\pi}{3} + \cos \frac{4\pi}{3} \sin \frac{13\pi}{6} = \frac{1}{2}$

$\displaystyle \text{(iv) } \sin \frac{10\pi}{3} \cos \frac{13\pi}{6} + \cos \frac{8\pi}{3} \sin \frac{5\pi}{6} = -1$

$\displaystyle \text{(v) } \tan \frac{5\pi}{4} \cot \frac{9\pi}{4} + \tan \frac{17\pi}{4} \cot \frac{15\pi}{4} = 0$

$\displaystyle \text{(i) } \text{LHS } = \tan 4\pi - \cos \frac{3\pi}{2} - \sin \frac{5\pi}{6} \cos \frac{2\pi}{3}$

$\displaystyle = 0 - 0 - \sin \Big( \pi - \frac{\pi}{6} \Big) \cos \Big( \frac{\pi}{2} + \frac{\pi}{6} \Big)$

$\displaystyle = - \sin \frac{\pi}{6} \Big( - \sin \frac{\pi}{6} \Big)$

$\displaystyle = \sin^2 \frac{\pi}{6} = \Big( \frac{1}{2} \Big)^2 = \frac{1}{4} = \text{ RHS. Hence proved. }$

$\displaystyle \text{(ii) } \text{LHS } = \sin \frac{13\pi}{3} \sin \frac{2\pi}{3} + \cos \frac{4\pi}{3} \sin \frac{13\pi}{6}$

$\displaystyle = \sin \Big( 4\pi + \frac{\pi}{3} \Big) \sin \Big( 3\pi - \frac{\pi}{3} \Big) + \cos \Big( \frac{\pi}{2} + \frac{\pi}{6} \Big) \sin \Big( \pi - \frac{\pi}{6} \Big)$

$\displaystyle = \sin \frac{\pi}{3} \sin \frac{\pi}{3} - \sin \frac{\pi}{6} \sin \frac{\pi}{6}$

$\displaystyle = \sin^2 \frac{\pi}{3} - \sin^2 \frac{\pi}{6}$

$\displaystyle = \Big( \frac{\sqrt{3}}{2} \Big)^2 - \Big( \frac{1}{2} \Big)^2 = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \text{(iii) } \text{LHS } = \sin \frac{13\pi}{3} \sin \frac{2\pi}{3} + \cos \frac{4\pi}{3} \sin \frac{13\pi}{6}$

$\displaystyle = \sin \Big( 4\pi + \frac{\pi}{3} \Big) \sin \Big( \pi - \frac{\pi}{3} \Big) + \cos \Big( \pi + \frac{\pi}{3} \Big) \sin \Big( 2\pi - \frac{\pi}{6} \Big)$

$\displaystyle = \sin \frac{\pi}{3} \sin \frac{\pi}{3} - \cos \frac{\pi}{3} \sin \frac{\pi}{6}$

$\displaystyle = \sin^2 \frac{\pi}{3} - \cos \frac{\pi}{3} \sin \frac{\pi}{6}$

$\displaystyle = \Big( \frac{\sqrt{3}}{2} \Big)^2 - \Big( \frac{1}{2} \Big)^2 = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \text{(iv) } \text{LHS } = \sin \frac{10\pi}{3} \cos \frac{13\pi}{6} + \cos \frac{8\pi}{3} \sin \frac{5\pi}{6}$

$\displaystyle = \sin \Big( 3\pi + \frac{\pi}{3} \Big) \cos \Big( 2\pi + \frac{\pi}{6} \Big) + \cos \Big( 3\pi - \frac{\pi}{3} \Big) \sin \Big( \pi - \frac{\pi}{6} \Big)$

$\displaystyle = \sin \Big( \pi + \frac{\pi}{3} \Big) \cos \Big( \frac{\pi}{6} \Big) + \cos \Big( \pi - \frac{\pi}{3} \Big) \sin \Big( \frac{\pi}{6} \Big)$

$\displaystyle = - \sin \frac{\pi}{3} \cos \frac{\pi}{6} - \cos \frac{\pi}{3} \sin \frac{\pi}{6}$

$\displaystyle = - \Big( \frac{\sqrt{3}}{2} \Big)^2 - \Big( \frac{1}{2} \Big)^2 = - \frac{3}{4} - \frac{1}{4} = -1 = \text{ RHS. Hence proved. }$

$\displaystyle \text{(v) } \text{LHS } = \tan \frac{5\pi}{4} \cot \frac{9\pi}{4} + \tan \frac{17\pi}{4} \cot \frac{15\pi}{4}$

$\displaystyle = \tan \Big( \pi + \frac{\pi}{4} \Big) \cot \Big( 2\pi + \frac{\pi}{4} \Big) + \tan \Big( 4\pi + \frac{\pi}{4} \Big) \cot \Big( 4\pi - \frac{\pi}{4} \Big)$

$\displaystyle = \tan \frac{\pi}{4} \cot \frac{\pi}{4} + \tan \frac{\pi}{4} \Big( - \cot \frac{\pi}{4} \Big)$

$\displaystyle = 1 -1 = 0 = \text{ RHS. Hence proved. }$