# RS Aggarwal Class 10 Solutions Chapter 8 - Trigonometric Identities - Ex 8A(8.1)

## RS Aggarwal Class 10 Chapter 8 - Trigonometric Identities - Ex 8A(8.1) Solutions Free PDF

All these RS Aggarwal class 10 solutions Chapter 8 Exercise 8.1 : Trigonometric Identities are solved by Byju's top ranked professors as per CBSE guidelines.

Q.1: Prove that:

(i) $sin \Theta cos ( 90^{\circ} – \Theta ) + sin ( 90^{\circ} – \Theta ) cos \Theta = 1$

Sol:

LHS = $sin \Theta cos ( 90^{\circ} – \Theta ) + sin ( 90^{\circ} – \Theta ) cos \Theta = 1$

= $sin \Theta sin \Theta + cos \Theta cos \Theta$ = $sin ^{ 2 } \Theta + cos ^{ 2 } \Theta$ = 1 = RHS

Hence Proved

(ii) $\frac{ sin \Theta } { cos ( 90^{\circ} – \Theta ) } + \frac{ cos \Theta }{ sin (90 ^{\circ} – \Theta ) } = 2$

Sol:

LHS = $\frac{ sin \Theta }{ cos ( 90 ^{\circ} – \Theta ) } + \frac{ cos \Theta }{ sin (90 ^{\circ} – \Theta ) }$

= $\frac{ sin \Theta }{ sin \Theta } + \frac{ cos \Theta }{ cos \Theta }$ = 1 + 1 = 2 = RHS

Hence Proved

(iii) $\frac{ sin \Theta \; cos ( 90 ^{\circ} – \Theta ) \; cos \Theta }{ sin ( 90 ^{\circ} – \Theta ) } + \frac{ cos \Theta \; sin( 90 ^{\circ} – \Theta) \; sin \Theta }{ cos (90 ^{\circ} – \Theta )} = 1$

Sol:

LHS = $\frac{ sin \Theta \; cos( 90 ^{\circ} – \Theta ) \; cos \Theta }{sin ( 90 ^{\circ} – \Theta ) } + \frac{ cos \Theta \; sin ( 90 ^{\circ} – \Theta) \; sin \Theta }{ cos ( 90 ^{\circ} – \Theta ) } = 1$

= $\frac{ sin \Theta sin \Theta cos \Theta }{cos \Theta } + \frac{ cos \Theta cos \Theta sin \Theta }{ sin \Theta }$ = $sin ^{ 2 } \Theta + cos^{ 2 } \Theta$ = 1 = RHS

Hence Proved

(iv) $\frac{ cos ( 90 ^{\circ}-\Theta) sec ( 90 ^{\circ} – \Theta ) tan \Theta }{cosec ( 90 ^{\circ} – \Theta ) sin ( 90 ^{\circ} – \Theta ) cot ( 90 ^{\circ} – \Theta )} + \frac{ tan ( 90 ^{\circ} – \Theta ) }{ cot \Theta } = 2$

Sol:

LHS = $\frac{ cos ( 90 ^{\circ} – \Theta ) sec ( 90 ^{\circ} – \Theta ) tan \Theta }{ cosec ( 90 ^{\circ} – \Theta ) sin ( 90^{\circ} – \Theta ) cot ( 90 ^{\circ} – \Theta ) } + \frac{ tan ( 90^{\circ} – \Theta ) }{ cot \Theta } = 2$

= $\frac{ sin \Theta cosec \Theta tan \Theta }{ sec \Theta cos \Theta tan \Theta } + \frac{ cot\Theta }{ cot \Theta }$ = 1 + 1 = 2 = RHS

Hence Proved

(v) $\frac{ cos ( 90 ^{\circ} – \Theta ) }{ 1 + sin ( 90^{\circ} – \Theta ) } + \frac{ 1 + sin( 90^{\circ} – \Theta ) }{ cos ( 90 ^{\circ} – \Theta ) } = 2 \; cosec \Theta$

Sol:

LHS = $\frac{ cos ( 90 ^{\circ} – \Theta ) }{ 1 + sin ( 90 ^{\circ} – \Theta ) } + \frac{ 1 + sin(90 ^{\circ} – \Theta )}{ cos ( 90^{\circ} – \Theta ) }$

= $\frac{ sin \Theta }{ 1 + cos \Theta } + \frac{ 1 + cos\Theta }{ sin \Theta }$

= $\frac{ sin ^{ 2 } \Theta + ( 1 + cos \Theta ) ^{ 2 } } { ( 1 + cos \Theta ) sin \Theta }$

= $\frac{ sin ^{ 2 } \Theta + 1 + cos ^{ 2 } \Theta + 2 \; cos \Theta } { ( 1 + cos \Theta )sin \Theta }$

= $\frac{ 1 + 1 + 2 \; cos \Theta } { ( 1 + cos \Theta ) sin \Theta }$

= $\frac{ 2 + 2 \; cos \Theta } { ( 1 + cos \Theta ) sin \Theta }$

= $\frac{2 ( 1 + cos \Theta ) } { ( 1 + cos \Theta ) sin \Theta }$ = $2 \frac{ 1 }{ sin \Theta }$ = $2 cosec \Theta$ = RHS

Hence Proved

??

(vi) $\frac{ sec ( 90 ^{\circ} – \Theta ) \; cosec \Theta – tan ( 90 ^{\circ} – \Theta ) \; cot \Theta + cos ^{2} 25 ^{\circ} + cos ^{2} 65 ^{\circ} }{ 3 \; tan 27 ^{\circ} \; tan 63 ^{\circ} } = \frac{ 2 }{ 3 }$

Sol:

LHS = $\frac{ sec ( 90^{\circ} – \Theta ) \; cosec \Theta – tan ( 90 ^{\circ} – \Theta ) \; cot \Theta + cos ^{2} 25 ^{\circ} + cos ^{2} 65 ^{\circ} }{ 3 \; tan 27 ^{\circ} \; tan 63 ^{\circ} }$

= $\frac{ cosec \Theta \; cosec \Theta – cot \Theta \; cot \Theta + sin ^{ 2 }( 90 ^{\circ} – 25 ^{\circ}) + cos ^{2} 65 ^{\circ} }{ 3 \; tan 27 ^{\circ} \; cot( 90^{\circ} – 63 ^{\circ} ) }$

= $\frac{ cosec ^{2} \Theta – cot ^{2} + sin ^{2} 65 ^{\circ} + cos ^{2} 65 ^{\circ} }{ 3 \; tan 27 ^{\circ} \; cot 27 ^{\circ} }$

= $\frac{ 1 + 1 }{ 3 \times tan 27^{\circ} \times \frac{ 1 }{ tan 27 ^{\circ} } }$ = $\frac{ 2 }{ 3 }$ = RHS

(vii) $cot \Theta \; tan ( 90 ^{\circ} – \Theta ) – sec ( 90 ^{\circ} – \Theta ) cosec \Theta + \sqrt{ 3 } tan 12 ^{\circ} tan 60 ^{\circ} tan 78 ^{\circ} = 2$

Sol:

LHS = ??$cot \Theta \; tan ( 90 ^{\circ} – \Theta ) – sec ( 90 ^{\circ} – \Theta ) cosec \Theta + \sqrt{ 3 } tan 12 ^{\circ} tan 60 ^{\circ} tan 78 ^{\circ}$

= $cot \Theta \; cot \Theta – cosec\Theta cosec \Theta + \sqrt{ 3 } tan 12 ^{\circ} \times \sqrt{ 3 } \times cot ( 90 ^{\circ} – 78 ^{\circ} )$

= $cot^{2} \Theta – cosec^{2} \Theta + 3 \; tan 12^{\circ} \; cot 12^{\circ}$

= -1 + 3 * $tan 12^{\circ} \times \frac{1}{tan 12^{\circ}}$ = -1 + 3 = 2 = RHS

??

Q.2: Prove that:

(i) $tan 5 ^{\circ} \; tan 25 ^{\circ} \; tan 30 ^{\circ} \; tan 65 ^{\circ} \; tan 85 ^{\circ} = \frac{ 1 }{\sqrt{ 3 } }$

Sol:

LHS = $tan 5 ^{\circ} \; tan 25 ^{\circ} \; tan 30 ^{\circ} \; tan 65 ^{\circ} \; tan 85 ^{\circ} = \frac{ 1 }{\sqrt{ 3 } }$

= $tan ( 90 ^{\circ} – 85 ^{\circ} ) \; tan ( 90 ^{\circ} – 65 ^{\circ}) \times \frac{ 1 }{ \sqrt{ 3 } } \times \frac{ 1 }{cot 65^{\circ}} \frac{ 1 }{ cot 85 ^{\circ}}$

= $cot 85^{\circ} \; cot 65^{\circ} \frac{ 1 }{ \sqrt{ 3 } } \; \frac{ 1 }{ cot 65 ^{\circ} } \frac{ 1 }{ cot 85 ^{\circ} }$ = $\frac{ 1 }{ \sqrt{ 3 } } = RHS$

(ii) $cot 12 ^{\circ} \; cot 38 ^{\circ} \; cot 52 ^{\circ} \; cot 60 ^{\circ} \; cot 78 ^{\circ} = \frac{ 1 }{\sqrt{ 3 } }$

Sol:

LHS = $cot 12 ^{\circ} \; cot 38 ^{\circ} \; cot 52 ^{\circ} \; cot 60 ^{\circ} \; cot 78 ^{\circ}$

= $tan ( 90 ^{\circ} – 12 ^{\circ} ) \times tan ( 90^{\circ} – 38 ^{\circ}) \times cot 52 ^{\circ} \times \frac{ 1 }{ \sqrt{ 3 } } \times cot 78 ^{\circ}$

= $\frac{ 1 }{\sqrt{ 3 } } \times tan 78 ^{\circ} \times tan 52 ^{\circ} \times cot 52 ^{\circ} \times cot 78 ^{\circ}$

= $\frac{ 1 } {\sqrt{ 3 } } \times tan 78^{\circ} \times tan 52^{\circ} \times \frac{ 1 }{tan 52 ^{\circ}} \times \frac{ 1 }{tan 78 ^{\circ}}$ = $\frac{ 1 }{\sqrt{ 3 } }$ = RHS

??

Sol:

$cos 15^{\circ} \; cos 35^{\circ} \; cosec 55^{\circ} \; ^{\circ}cos 60^{\circ} \; cosec 75^{\circ} = \frac{ 1 }{ 2 }$

(iii) LHS = $cos 15^{\circ} \; cos 35^{\circ} \; cosec 55^{\circ} \; ^{\circ}cos 60^{\circ} \; cosec 75^{\circ}$

= $cos ( 90 ^{\circ} – 75 ^{\circ} ) cos ( 90 ^{\circ} – 55 ^{\circ}) \; \frac{ 1 }{ sin 55 ^{\circ} } \times \frac{ 1 }{ 2 } \times \frac{ 1 }{ sin 75 ^{\circ}}$

= $sin 75 ^{\circ} sin 55 ^{\circ} \frac{ 1 }{ sin 55^{\circ}} \times \frac{ 1 }{2 } \times \frac{ 1 }{ sin 75^{\circ} }$ = $\frac{ 1 }{ 2 }$ = RHS

(iv) $cos 1 ^{\circ} \; cos 2 ^{\circ} \; cos 3 ^{\circ} … cos 180 ^{\circ} = 0$

Sol:

LHS = $cos 1^{\circ} \; cos 2^{\circ} \; cos 3^{\circ} … cos 180^{\circ}$

= $cos 1^{\circ} \times cos 2^{\circ} \times cos 3^{\circ} \times …\times cos 90^{\circ} \times …cos 180^{\circ}$

= $cos 1^{\circ} \times cos 2^{\circ} \times cos 3^{\circ} \times …\times 0 \times …cos 180^{\circ}$ = 0 = RHS

(v) $\left (\frac{ sin 49 ^{\circ} }{ cos 41 ^{\circ}} \right ) ^{ 2 } + \left ( \frac{ cos 41^{\circ} }{ sin 49 ^{\circ} } \right ) ^{ 2 } = 2$

Sol:

LHS = $\left (\frac{ sin 49^{\circ} }{ cos 41^{\circ}} \right ) ^{ 2 } + \left ( \frac{ cos 41 ^{\circ} }{ sin 49^{\circ}} \right ) ^{ 2 }$

= $\left (\frac{ cos (90 ^{\circ} – 49 ^{\circ} ) }{ cos 41 ^{\circ} } \right ) ^{2} + \left ( \frac{ cos 41^{\circ} }{ cos ( 90 ^{\circ} – 49 ) } \right ) ^{ 2 }$

= $\left (\frac{ cos 41 ^{\circ} }{ cos 41^{\circ}} \right ) ^{2} + \left ( \frac{ cos 41^{\circ} }{cos 41 ^{\circ} } \right ) ^{ 2 }$

= $1^{ 2 } + 1^{ 2 }$ = 1 + 1 = 2 = RHS

??

Q.3: Prove that:

(i) $sin ( 70 ^{\circ} + \Theta ) – cos ( 20^{\circ} – \Theta ) = 0$

Sol:

LHS = $sin( 70 ^{\circ} + \Theta ) – cos( 20 ^{\circ} – \Theta )$

= $sin\left \{ 90 ^{\circ} – ( 20 ^{\circ} – \Theta ) \right \} – cos ( 20 ^{\circ} – \Theta )$

= $cos ( 20 ^{\circ} – \Theta ) – cos ( 20^{\circ} – \Theta )$ = 0 = RHS

??

(ii) $tan ( 55^{\circ} – \Theta ) – cot ( 35^{\circ} + \Theta ) = 0$

Sol:

LHS = $tan (55^{\circ} – \Theta ) – cot (35^{\circ} + \Theta )$

= $\tan \left [ 90 ^{\circ} – \left ( 35 ^{\circ} + \Theta \right ) \right ] – \left ( \cot 35^{\circ} + \Theta \right )$

= $cot ( 35^{\circ} + \Theta ) – cot( 35^{\circ} + \Theta )$ = 0 = R.H.S.

(iii) $cosec ( 67 ^{\circ} + \Theta ) – sec ( 23^{\circ} – \Theta ) = 0$

Sol:

LHS = $cosec ( 67^{\circ} + \Theta ) – sec ( 23^{\circ} – \Theta )$

= $cosec \left \{ 90 ^{\circ} – ( 23 ^{\circ} – \Theta ) \right \} – sec( 23^{\circ} – \Theta )$

= $sec ( 23^{\circ} – \Theta ) – sec ( 23^{\circ} – \Theta )$ = 0 = RHS

(iv) $cosec (65 ^{\circ} + \Theta ) – sec ( 25^{\circ} – \Theta )- tan ( 55^{\circ} – \Theta ) + cot ( 35 ^{\circ} + \Theta ) = 0$

Sol:

$cosec ( 65^{\circ} + \Theta ) – sec( 25 ^{\circ} – \Theta )- tan ( 55^{\circ} – \Theta ) + cot( 35^{\circ} + \Theta )$

= $cosec \left \{ 90 ^{\circ} – ( 25 ^{\circ} – \Theta ) \right \} – sec ( 25 ^{\circ} – \Theta ) – tan ( 55 ^{\circ} – \Theta ) + cot \left \{ 90^{\circ} – (55^{\circ} – \Theta ) \right \}$

= $sec (25^{\circ} – \Theta ) – sec (25^{\circ} – \Theta ) – tan (55^{\circ} – \Theta ) + tan (55^{\circ} – \Theta )$ = 0 = RHS

(v) $sin ( 50 ^{\circ} + \Theta ) – cos ( 40 ^{\circ} – \Theta ) + tan 1 ^{\circ} \; tan 10 ^{\circ} \; tan 80 ^{\circ} \; tan 89 ^{\circ} = 1$

Sol:

LHS = $sin ( 50 ^{\circ} + \Theta ) – cos ( 40 ^{\circ} – \Theta ) + tan 1 ^{\circ} \; tan 10 ^{\circ} \; tan 80 ^{\circ} \; tan 89 ^{\circ}$

= $sin \left \{ 90 ^{\circ} – ( 40 ^{\circ} – \Theta ) \right \} – cos ( 40^{\circ} – \Theta ) \; + \; \left \{ tan 1 ^{\circ} tan ( 90 ^{\circ} – 1 ^{\circ} ) \right \}\left \{ tan 10 ^{\circ} tan ( 90 ^{\circ} – 10 ^{\circ}) \right \}$

= $cos ( 40 ^{\circ} – \Theta ) – cos ( 40 ^{\circ} – \Theta ) + (tan 1^{\circ} cot 1 ^{\circ}) (tan 10 ^{\circ} cot 10 ^{\circ} )$

= $\left ( \frac{ 1 }{cot 1 ^{\circ} } \times cot \; 1 ^{\circ} \right )\left ( tan 10 ^{\circ} \times \frac{ 1 }{ tan 10^{\circ} } \right )$ = 1 * 1 = 1= RHS

??

Q.8: Express each of the following in terms of T-ratios of angles lying between $0^{\circ}$ and $45^{\circ}$

(i) $sin 67 ^{\circ} + cos 75 ^{\circ}$

Sol:

$sin 67 ^{\circ} + cos 75 ^{\circ}$

= $cos ( 90^{\circ} – 67 ^{\circ} ) + sin ( 90 ^{\circ} – 75 ^{\circ} )$

= $cos 23 ^{\circ} + sin 15 ^{\circ}$

(ii) $cot 65 ^{\circ} + tan 49 ^{\circ}$

Sol:

$cot 65^{\circ} + tan 49^{\circ}$

= $tan (90^{\circ} – 65^{\circ}) + cot (90^{\circ} – 49^{\circ})$

= $tan 25^{\circ} \; + \; cot 41^{\circ}$

(iii) $sec 78^{\circ} \; + \; cosec 56^{\circ}$

Sol:

$sec 78^{\circ} \; + \; cosec 56^{\circ}$

= $sec (90^{\circ} – 12^{\circ}) \; + \; cosec (90^{\circ} – 34^{\circ})$

= $cosec 12^{\circ} \; + \; sec 34^{\circ}$

(iv) $cosec 54^{\circ} \; + \; sin 72^{\circ}$

Sol:

$cosec 54^{\circ} \; + \; sin 72^{\circ}$

= $sec (90^{\circ} – 54^{\circ}) \; + \; cos(90^{\circ} – 72^{\circ})$

= $sec 36^{ \circ} \; ??+ ??\; cos18^{ \circ}$

??

Q.4: If A, B and C are the angles of a $\triangle ABC$ , prove that $tan \left (\frac{ C + A }{ 2 } \right )$ = $cot \frac{ B }{ 2 }$

Sol:

In $\triangle ABC$

A + B + C = $180^{ \circ }$ $\Rightarrow A + C = 180 ^{\circ} – B$ . . . . . . . . . ??. . (i)

Now, LHS = $tan \left (\frac{ C + A }{ 2 } \right )$

= $tan \left ( \frac{ 180^{\circ} – B }{ 2 } \right )$ [Using (i)]

= $tan \left ( 90^{\circ} – \frac{ B }{ 2 } \right )$ = $cot \frac{ B }{ 2 }$ = RHS

Q.5: If $cos \; 2 \Theta = sin \; 4 \Theta$ , where $2 \Theta$ and $4 \Theta$ are acute angles , find the value of $\Theta$ .

Sol:

We have, $cos \; 2 \Theta = sin \; 4 \Theta$ $\Rightarrow sin (90^{\circ} – 2 \Theta ) = sin 4 \Theta$

Comparing both sides, we get:

$90 ^{ \circ } – 2 \Theta = 4 \Theta$ $\Rightarrow 2 \Theta + 4 \Theta = 90 ^{\circ}$ $\Rightarrow 6 \Theta = 90 ^{\circ}$ $\Rightarrow \Theta = \frac{ 90 ^{\circ} }{ 6 }$ $Therefore, \Theta = 15 ^{\circ}$

Hence, the value of $\Theta$ is $15 ^{\circ}$

Q.6: If $sec 2A = cosec ( A – 42^{\circ} )$ , where 2A is an acute angle , find the value of A.

Sol:

We have, $sec 2A = cosec ( A – 42 ^{\circ} )$ $\Rightarrow cosec ( 90 ^{\circ} – 2A ) = cosec ( A – 42^{\circ} )$

Comparing both sides, we get

$90 ^{\circ} – 2A = A – 42 ^{\circ}$ $\Rightarrow 2A + A = 90 ^{\circ} + 42 ^{\circ}$ $\Rightarrow 3A = 132 ^{\circ}$ $\Rightarrow A = \frac{ 132 ^{\circ} }{ 3 }$ $Therefore, A = 44 ^{\circ}$

Hence, the value of A is $44 ^{\circ}$

Q.7: If $sin 3 A = cos \left ( A – 26 ^{\circ} \right )$ , where 3A is an acute angle, find the value of A.

Sol:

$sin 3 A = cos \left ( A – 26 ^{\circ} \right )$ $\Rightarrow cos ( 90 ^{\circ} – 3A ) = cos (A – 26 ^{\circ})$ [ $Since, sin \Theta = cos ( 90 ^{\circ} – \Theta )$ ] $\Rightarrow 90 ^{\circ} – 3A = A – 26 ^{\circ}$ $\Rightarrow 116^{\circ} = 4 A$ $\Rightarrow A = \frac{ 116 ^{\circ} }{ 4 } = 29 ^{\circ}$

Q.8: If $tan \; 2A = cot ( A – 12 ^{\circ})$ , where 2A is an acute angle , find the value of A.

Sol:

$tan \; 2A = cot ( A – 12 ^{\circ})$ $\Rightarrow cot ( 90^{\circ} – 2A ) = cot ( A – 12 ^{\circ} )$ [ $Since, tan \Theta = cot ( 90 ^{\circ} -\Theta )$] $\Rightarrow ( 90 ^{\circ} – 2A ) = ( A – 12 ^{\circ})$ $\Rightarrow 102 ^{\circ} = 3A$ $\Rightarrow A = \frac{ 102 ^{\circ}}{ 3 } = 34 ^{\circ}$

Q.9: If $sec \; 4A = cosec ( A – 15 ^{\circ} )$ , where 4A is an acute angle, find the value of A.

Sol:

??$sec \; 4A = cosec ( A – 15 ^{\circ} )$ $\Rightarrow cosec ( 90 ^{\circ} – 4A ) = cosec ( A – 15 ^{\circ} )$ [ $Since, sec \Theta = cosec ( 90 ^{\circ} – \Theta )$ ] $\Rightarrow 90 ^{\circ} – 4A = A – 15 ^{\circ}$ $\Rightarrow 105 ^{\circ} = 5A$ $\Rightarrow A = \frac{ 105 ^{\circ} }{ 5 } = 21 ^{\circ}$

??

Q.10: Prove that:

$\frac{ 2 }{ 3 } cosec ^{2} \; 58^{\circ} – \frac{ 2 }{ 3 } cot 58 ^{\circ} \; tan 32 ^{\circ} – \frac{ 5 }{ 3 } tan 13 ^{\circ} \; tan 37 ^{\circ} \; tan 45 ^{\circ} \; tan 53 ^{\circ} \; tan 77 ^{\circ} = -1$

Sol: $\frac{ 2 }{ 3 } cosec ^{2} \; 58^{\circ} – \frac{ 2 }{ 3 } cot 58 ^{\circ} \; tan 32 ^{\circ} – \frac{ 5 }{ 3 } tan 13 ^{\circ} \; tan 37 ^{\circ} \; tan 45 ^{\circ} \; tan 53 ^{\circ} \; tan 77 ^{\circ}$

= $\frac{ 2 }{ 3 } \left (cosec ^{2} \; 58^{\circ} – cot 58 ^{\circ} \; tan 32 ^{\circ} \right ) \; – \; \frac{ 5 }{ 3 } tan 13 ^{\circ} \; tan ( 90 ^{\circ} – 13 ^{\circ}) \; tan 37 ^{\circ} \; tan ( 90 ^{\circ} – 37 ^{\circ}) \; (tan 45 ^{\circ})$

= $\frac{ 2 }{ 3 } \left (cosec ^{2} \; 58^{\circ} – cot 58 ^{\circ} \; tan ( 90 ^{\circ} – 58 ^{\circ}) \right ) \; – \; \frac{ 5 }{ 3 } tan 13 ^{\circ} \; cot 13 ^{\circ} \; tan 37 ^{\circ} \; cot37 ^{\circ} \; (1)$

= $\frac{ 2 }{ 3 } \left (cosec ^{2} \; 58^{\circ} – cot 58 ^{\circ} \; cot 58 ^{\circ} \right ) \; – \; \frac{ 5 }{ 3 } tan 13 ^{\circ} \; \frac{ 1 }{ tan 13^{\circ} } \; tan 37 ^{\circ} \; \frac{ 1 }{tan 37 ^{\circ} }$

= $\frac{ 2 }{ 3 } \left ( cosec ^{2} 58^{\circ} – cot ^{2} 58 ^{\circ} \right ) \; – \;\frac{ 5 }{ 3 }$ = $\frac{ 2 }{ 3 } \; – \; \frac{ 5 }{ 3 }$ = -1

Hence Proved

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