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

i) \sin \Big( \frac{5 \pi}{3} \Big)        ii) \sin 17 \pi        iii) \tan \Big( \frac{11 \pi}{6} \Big)        iv) \cos \Big( - \frac{25\pi}{4} \Big)        v) \tan \Big( \frac{7\pi}{4} \Big)  

vi) \sin \Big( \frac{17\pi}{6} \Big)        vii) \cos \Big( \frac{19\pi}{6} \Big)        viii) \sin \Big(  - \frac{11\pi}{6} \Big)        ix) \mathrm{cosec} \Big(  - \frac{20\pi}{3} \Big)    

x) \tan \Big(  - \frac{13\pi}{4} \Big)        xi) \cos \Big( \frac{19\pi}{4} \Big)        xii)  \sin \Big( \frac{41\pi}{4} \Big)        xiii) \cos \Big( \frac{39\pi}{4} \Big)

xiv) \sin \Big( \frac{151\pi}{6} \Big)  

Answer:

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

ii)   \sin 17 \pi = 0 (\because n\pi = 0 for all n \in Z)

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

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: \cos(-x) =  \cos x ]

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

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

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

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}

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}}

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

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}

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

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

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}

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Question 2: Prove that:

i) \tan 225^o  \cot 405^o + \tan 765^o \cot 675^o = 0

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

iii) \cos 24^o + \cos 55^o + \cos 125^o + \cos 204^o + \cos 300^o = \frac{1}{2} 

iv) \tan (-225^o) \cot(-405^o) - \tan (-765^o) \cot (675^o) = 0

v) \cos 570^o \sin 510^o + \sin (-300^o) \cos (-390^o) = 0

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} 

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

Answer:

i) LHS = \tan 225^o  \cot 405^o + \tan 765^o \cot 675^o

= \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)

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

= 1 - 1 = 0 = RHS. Hence proved.

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

= \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)

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

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

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

iii) LHS = \cos 24^o + \cos 55^o + \cos 125^o + \cos 204^o + \cos 300^o

= ( \cos 24^o + \cos 204^o) + ( \cos 55^o + \cos 125^o) + \cos 300^o

= ( \cos 24^o + \cos (\pi + 24^o) ) + ( \cos 55^o + \cos (\pi - 55^o)) + \cos (2\pi - 60^o)

= \cos 24^o - \cos 24^o + \cos 55^o - \cos 55^o + \cos 60^o

= \cos 60^o = \frac{1}{2} = RHS. Hence proved.

iv) LHS = \tan (-225^o) \cot(-405^o) - \tan (-765^o) \cot (675^o)

= -\tan 225^o (-\cot 405^o ) + \tan 765^o \cot 675^o

= \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)

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

= 1 - 1 = 0 = RHS. Hence proved.

v)  LHS = \cos 570^o \sin 510^o + \sin (-300^o) \cos (-390^o)

= \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)

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

= 0 = RHS. Hence proved.

vi) 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}

= \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)

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

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

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

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

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

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

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

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Question 3: Prove that:

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

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

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

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

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

Answer:

i) 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) }

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

= 1 = RHS

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

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

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

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

= 1 + 1 = 2 = RHS. Hence proved.

iii) 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 ) }

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

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

= \tan^2 x \cot^2 x

= 1= RHS. Hence proved.

iv) 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\}

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

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

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

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

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

= 2 \cot x = RHS. Hence proved.

v) 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 )}

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

= 1 = RHS. Hence proved.

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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

Answer:

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

= \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}

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

= 1 + 1 = 2 = RHS. Hence proved.

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Question 5: Prove that:

\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

Answer:

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)

= \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)

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

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

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

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

= - 1 = RHS. Hence proved.

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

ii) \cos \Big( \frac{A+B}{2} \Big) = \sin \Big( \frac{C}{2} \Big)    iii) \tan \Big( \frac{A+B}{2} \Big) = \cot \Big( \frac{C}{2} \Big)

Answer:

i) LHS = \cos (A+B) + \cos C

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

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

= 0 =  RHS. Hence proved.

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

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

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

Answer:

Given A + C = 180^o, B+D = 180^o

LHS = \cos (180^o - A) + \cos (180^o + B) + \cos (180^o + C ) - \sin (90^o + D)

= \cos C + \cos D - \cos C - \cos D = 0 = RHS. Hence proved.

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Question 8: Find x from the following equations:

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)

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

Answer:

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)

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

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

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

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

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

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

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

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

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

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Question 9: Prove that:

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

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

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

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

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

Answer:

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

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

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

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

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

= \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)

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

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

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

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

= \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)

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

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

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

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

= \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)

= \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)

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

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

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

= \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)

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

= 1 -1 = 0 = RHS. Hence proved.

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