# RS Aggarwal Class 10 Solutions Chapter 7 Trigonometric Ratios Of Complementary Angles

## RS Aggarwal Class 10 Chapter 7 Trigonometric Ratios Of Complementary Angles Solutions Free PDF

The complementary angles are simply the angles, which when combined have a sum which must be equal to 90 degrees. Considering a right angle triangle, the sum of angles in a triangle is 90 degrees which are a measure of a right angle. There is a relationship between an acute angle and the length of a few sides of a right angle triangle.

The different trigonometric ratios of complementary angles are:

1. Sin (90- A)
2. Cos (90- A)
3. Tan (90- A)
4. Cosec (90- A)
5. Sec (90- A)
6. Cot (90- A)

## Download PDF of RS Aggarwal Class 10 Chapter 7- Trigonometric Ratios of Complementary Angles

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Q.1: Without using trigonometric table, evaluate:

(i) $\frac{ sin 16^{\circ} }{ cos 74^{\circ} }$

= $\frac{ sin ( 90^{\circ} – 74^{\circ}) }{ cos 74^{\circ} }$

= $\frac{ cos 74^{\circ}}{ cos 74^{\circ} }$ [ $Since, sin( 90-\Theta ) = cos\Theta$ ] = 1

(ii) $\frac{sec \; 11^{\circ}}{cosec \; 79^{\circ}}$

= $\frac{sec ( 90^{\circ} – 79^{\circ})}{cosec\;79^{\circ}}$

= $\frac{cosec \; 79^{\circ}}{cosec \; 79^{\circ}}$ [$Since, sec( 90 – \Theta )=cosec \Theta$ ] = 1

(iii) $\frac{ tan \; 27 ^{\circ} }{ cot \; 63 ^{\circ} }$

= $\frac{ tan( 90 ^{\circ} – 63^{\circ}) }{ cot \; 63 ^{\circ} }$

= $\frac{ cot \; 63^{\circ} }{ cot \; 63 ^{\circ} }$ [ $Since, sin(90-\Theta ) = cot\Theta$ ] = 1

(iv) $\frac{ cos \; 35^{\circ} }{ sin \; 55 ^{\circ} }$

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

= $\frac{ sin \; 55^{\circ} }{ sin \; 55^{\circ} }$ [ $sin( 90 – \Theta ) = cos \Theta$ ] = 1

(v) $\frac{ cosec \; 42 ^{\circ} }{ sec \; 48 ^{\circ} }$

= $\frac{ cosec( 90 ^{\circ} – 48 ^{\circ} ) }{ sec \; 48 ^{\circ} }$ [ $Since, sec( 90-\Theta )=cosec\Theta$ ] = 1

(vi) $\frac{cot \; 38^{\circ}}{tan \; 52^{\circ}}$

= $\frac{cot( 90^{\circ} – 52^{\circ})}{tan \; 52^{\circ}}$

= $\frac{tan \; 52^{\circ}}{tan \; 52^{\circ}}$ [$Since, tan( 90 – \Theta ) = cot \Theta$ ] = 1

Q.2: Without using trigonometric tables, proves that:

(i) $cos \; 81 ^{\circ} – sin \; 9 ^{\circ} = 0$

LHS = $cos \; 81 ^{\circ} – sin \; 9 ^{\circ}$

= $cos ( 90 ^{\circ} – 9^{\circ} ) – sin 9 ^{\circ}$

= $sin 9 ^{\circ} – sin 9 ^{\circ}$ = 0

(ii) $tan 71 ^{\circ} – cot 19 ^{\circ} = 0$

LHS = $tan 71 ^{\circ} – cot 19 ^{\circ}$

= $tan( 90^{\circ} – 19^{\circ} ) – cot( 19 ^{\circ} )$

= $cot(19 ^{\circ} ) – cot ( 19 ^{\circ} )$ = 0

= RHS

(iii) $cosec ( 80 ^{\circ} ) – sec ( 10 ^{\circ}) = 0$

LHS = $cosec ( 80^{\circ} ) – sec( 10 ^{\circ} )$

= $cosec( 90 ^{\circ} – 10 ^{\circ}) – sec( 10 ^{\circ} )$

= $sec ( 10 ^{\circ} ) – sec ( 10^{\circ} )$ = 0 = RHS

(iv) $cosec ^{2} ( 72 ^{\circ} ) – tan ^{2} ( 18 ^{\circ}) = 1$

LHS = $cosec^{2} ( 72 ^{\circ} ) – tan ^{2}( 18^{\circ} )$

= $cosec ^{2} ( 90 ^{\circ} – 18 ^{\circ} ) – tan ^{2} ( 18 ^{\circ} )$

= $sec ^{2} ( 18^{\circ} ) – tan ^{2}( 18^{\circ} )$ = 1 = RHS

(v) $cos ^{2} ( 75 ^{\circ} ) – cos ^{2}( 15 ^{\circ} ) = 1$

LHS = $cos ^{2} ( 75 ^{\circ} ) – cos ^{2} ( 15 ^{\circ} )$

= $cos ^{2}( 90 ^{\circ} – 15^{\circ}) – cos ^{2} ( 15 ^{\circ} )$

= $sin ^{2} ( 15 ^{\circ} ) + cos ^{2} ( 15 ^{\circ} )$ = 1 = RHS

(vi) $tan ^{2} ( 66 ^{\circ} ) – cot ^{2} ( 24 ^{\circ} ) = 0$

LHS = $tan^{2}(66^{\circ}) – cot^{2}( 24 ^{\circ} )$

= $cot^{2}(24^{\circ})-cot^{2}(24^{\circ})$ = 0 = RHS

(vii) $sin^{ 2 }( 48^{\circ} ) + sin ^{2} ( 42^{\circ}) = 1$

LHS = $sin ^{2}(48 ^{\circ} ) + sin ^{2}( 42 ^{\circ} )$

= $sin ^{2} ( 90^{\circ} – 42 ^{\circ} ) + sin ^{2}(42 ^{\circ})$

= $cos ^{2} ( 42 ^{\circ} ) +sin ^{2} ( 42^{\circ} )$ = 1 = RHS

(viii) $cos ^{2} ( 57 ^{\circ} ) – sin ^{2} ( 33 ^{\circ} ) = 0$

LHS = $cos ^{2} ( 57 ^{\circ} ) – sin ^{2} ( 33 ^{\circ} )$

= $cos ^{2} ( 90 ^{\circ} – 33 ^{\circ} ) – sin ^{2} ( 33 ^{\circ} )$

= $sin ^{2} ( 33 ^{\circ} ) – sin ^{2} ( 33 ^{\circ} )$ = 0 = RHS

(ix) $( sin 65 ^{\circ} + cos 25 ^{\circ} )( sin 65 ^{\circ} – cos 25 ^{\circ}) = 0$

LHS = $( sin 65 ^{\circ} + cos 25 ^{\circ} ) ( sin 65 ^{\circ} – cos 25 ^{\circ} )$

= $sin ^{2} ( 65 ^ {\circ} ) – cos ^{2}(25^{\circ})$

= $sin ^{2} ( 90 ^{\circ} – 25 ^{\circ} ) – cos ^{2} ( 25 ^{\circ} )$

= $cos ^{2} ( 25 ^{\circ} ) – cos ^{2} ( 25 ^{\circ} )$

= $cos ^{2} ( 65 ^{\circ} ) – cos^{2} ( 25 ^{\circ} )$ = 0 = RHS

Q.3: Without using trigonometric tables, prove that:

(i) $sin 53 ^{\circ} cos 37 ^{\circ} + cos 53 ^{\circ} sin 37 ^{\circ} = 1$

LHS = $sin 53 ^{\circ} cos 37 ^{\circ} + cos 53 ^{\circ} sin 37 ^{\circ}$

= $sin ( 90 ^{\circ} – 37 ^{\circ} ) cos 37 ^{\circ} + cos ( 90 ^{\circ} -37^{\circ})sin37^{\circ}$

= $cos 37 ^{\circ} cos 37 ^{\circ} + sin 37 ^{\circ} sin 37^{\circ}$

= $cos ^{2} 37 ^{\circ} + sin ^{2} 37 ^{\circ}$ = 1 = RHS

(ii) $cos 54 ^{\circ} cos 36 ^{\circ} – sin 54 ^{\circ} sin 36^ {\circ} = 0$

LHS = $cos 54 ^{\circ} cos 36 ^{\circ} – sin 54 ^{\circ}sin 36 ^{\circ}$

= $cos ( 90 ^{\circ} – 36 ^{\circ} ) cos 36 ^{\circ} – sin ( 90 ^{\circ} -36^{\circ} )sin 36 ^{\circ}$

= $sin 36 ^{\circ} cos 36^{\circ} – cos 36 ^{\circ} sin 36 ^{\circ}$ = 0 = RHS

(iii) $sec 70 ^{\circ} sin 20 ^{\circ} + cos 20 ^{\circ} cosec 70 ^{\circ} = 2$

LHS = $sec 70^{\circ} sin 20 ^{\circ} + cos 20 ^{\circ} cosec 70 ^{\circ}$

= $sec ( 90 ^{\circ} – 20 ^{\circ} ) sin 20 ^{\circ} + cos 20 ^{\circ} cosec ( 90 ^{\circ} – 20 ^{\circ})$

= $cosec 20 ^{\circ} \frac{ 1 } { cosec 20 ^{\circ} } + \frac{ 1 }{ sec 20 ^{\circ} } sec 20 ^{\circ}$ = 1 + 1 = 2 = RHS

(iv) $sin 35 ^{\circ} sin 55 ^{\circ} – cos 35 ^{\circ} cos 55 ^{\circ} = 0$

LHS = $sin 35^ {\circ} sin 55 ^{\circ} – cos 35 ^{\circ} cos 55 ^{\circ}$

= $sin 35^{\circ} cos ( 90 ^{\circ} – 55^{\circ} ) – cos 35 ^{\circ} sin ( 90 ^{\circ} -55^{\circ} )$

= $sin 35 ^{\circ} cos 35 ^{\circ} – cos 35 ^{\circ} sin 35 ^{\circ}$ = 0 = RHS

(v) $(sin 72 ^{\circ} + cos 18 ^{\circ} ) ( sin 72 ^{\circ} – cos 18 ^{\circ} ) = 0$

LHS = $(sin 72 ^{\circ} + cos 18 ^{\circ} ) ( sin 72 ^{\circ} – cos 18 ^{\circ} )$

= $( sin 72 ^{\circ} + cos 18 ^{\circ} ) ( cos ( 90 ^{\circ} – 72 ^{\circ})-cos 18 ^{\circ} )$

= $( sin 72 ^{\circ} + cos 18 ^{\circ} ) ( cos 18 ^{\circ} – cos 18 ^{\circ} )$

= $(sin 72 ^{\circ} + cos 18 ^{\circ} ) ( 0 )$ = 0 = RHS

(vi) $tan 48 ^{\circ} tan 23^{\circ} tan 42 ^{\circ} tan 67 ^{\circ} = 1$

LHS = $tan 48 ^{\circ} tan 23 ^{\circ} tan 42 ^{\circ} tan 67 ^{\circ}$

= $cot ( 90^{\circ} – 48 ^{\circ} ) cot ( 90 ^{\circ} -23 ^{\circ} ) tan 42 ^{\circ} tan 67 ^{\circ}$

= $cot 42 ^{\circ} cot 67 ^{\circ} tan 42 ^{\circ} tan 67 ^{\circ}$

= $\frac{ 1 } { tan 42 ^{\circ} } \times \frac{ 1 }{ tan 67 ^{\circ}} \times tan 42 ^{\circ} \times tan 67 ^{\circ}$ = 1 = RHS

Q.4: Prove that:

(i) $\frac{ sin 70 ^{\circ} } { cos 20 ^{\circ} } + \frac{ cosec20 ^{\circ} } { sec 70 ^{\circ} } – 2 cos 70 ^{\circ} cosec 20 ^{\circ} = 0$

LHS = $\frac{ sin 70 ^ {\circ} } { cos 20 ^{\circ} } + \frac{ cosec 20 ^{\circ} }{ sec 70 ^{\circ} } – 2 cos 70 ^{\circ} cosec 20 ^{\circ}$

= $\frac{ sin 70 ^{\circ} }{ sin ( 90 ^{\circ} – 20 ^{\circ} ) } + \frac{ sec ( 90 ^{\circ} -20 ^{\circ} ) } { sec 70 ^{\circ} } – 2 cos 70 ^{\circ} sec ( 90 ^{\circ} – 20 ^{\circ} )$

= $\frac{ sin 70 ^{\circ} } { sin 70 ^{\circ} } + \frac{ sec 70 ^{\circ} }{ sec 70 ^{\circ} } -2 cos 70 ^{\circ} sec 70 ^{\circ}$

= 1 + 1 – 2 * $cos 70 ^{\circ} \times \frac{ 1 }{ cos 70 ^{\circ} }$ = 2 – 2 = 0 = RHS

(ii) $\frac{ cos 80 ^{\circ} }{ sin 10 ^{\circ} } + cos 59 ^{\circ} cosec 31^{\circ} = 2$

LHS = $\frac{ cos 80 ^{\circ} }{ sin 10 ^{\circ} } + cos 59 ^{\circ} cosec 31 ^{\circ}$

= $\frac{ cos 80^{\circ} }{ cos ( 90^{\circ} – 10^{\circ})}+sin( 90^{\circ} – 59^{\circ})cosec 31^{\circ}$

= $\frac{cos80^{\circ}}{cos80^{\circ}}+sin31^{\circ}\;cosec31^{\circ}$

= 1 + $sin31^{\circ}\times \frac{1}{sin31^{\circ}}$ = 1 + 1 = 2 = RHS

(iii) $\frac{ 2 sin 68 ^{\circ} }{ cos 22 ^{\circ} } – \frac{ 2 cot 15 ^{\circ} }{ 5 tan 75 ^{\circ} } – \frac{ 3 \; tan 45 ^{\circ} \; tan 20 ^{\circ} \; tan 40 ^{\circ} \; tan 50 ^{\circ} \; tan 70 ^{\circ} } { 5 }$

LHS = $\frac{ 2 sin 68 ^{\circ} } { cos 22 ^{\circ} } -\frac{ 2 cot 15 ^{\circ} } { 5 tan 75 ^{\circ}} -\frac{ 3 \; tan 45 ^{\circ} \; tan 20 ^{\circ} \; tan 40 ^{\circ} \; tan 50 ^{\circ} \; tan 70 ^{\circ} }{ 5 } = 1$

= $\frac{ 2 sin 68 ^{\circ} } { sin ( 90 ^{\circ} – 22 ^{\circ} ) } -\frac{ 2 cot 15 ^{\circ} }{ 5 cot ( 90 ^{\circ} – 75 ^{\circ} ) } – \frac{ 3 \times 1 \times cot ( 90 ^{\circ} – 20 ^{\circ} ) \times cot ( 90 ^{\circ} – 40 ^{\circ} ) \times tan 50 ^{\circ} \times tan 70 ^{\circ} }{ 5 }$

= $\frac{ 2 \; sin 68 ^{\circ} }{ sin 68 ^{\circ} } – \frac{ 2 cot 15 ^{\circ} }{ 5 cot 15 ^{\circ} } – \frac{ 3 \; cot 70 ^{\circ} \; cot 50 ^{\circ} \; tan 50 ^{\circ} \; tan 70 ^{\circ} }{ 5 }$

= $2 – \frac{ 2 } { 5 } – \frac{ 3 \times \frac{ 1 }{ tan 70 ^{\circ}}\times \frac{ 1 }{ tan 50 ^{\circ} } \times tan 50 ^{\circ} \times tan 70 ^{\circ} }{ 5 }$

= $2-\frac{ 2 } { 5 } -\frac{ 3 }{ 5 }$

= $\frac{10 – 2 – 3}{ 5 }$ = $\frac{ 5 }{ 5 }$ = 1 = RHS

(iv) $\frac{sin 18 ^{\circ} }{ cos 72 ^{\circ} } + \sqrt{ 3 }( tan 10 ^{\circ} \; tan 30 ^{\circ} \; tan 40 ^{\circ} \; tan 50 ^{\circ} ) = 2$

LHS = $\frac{ sin 18 ^{\circ} }{ cos 72 ^{\circ} } + \sqrt{ 3 }( tan 10 ^{\circ} \; tan 30 ^{\circ} \; tan 40 ^{\circ} \; tan 50 ^{\circ} )$

= $\frac{sin 18 ^{\circ} }{ sin ( 90 ^{\circ} – 72 ^{\circ})} + \sqrt{3} [ cot( 90^{\circ}-10 ^{\circ}) \times \frac{ 1 }{ \sqrt{ 3 } } \times cot ( 90 ^{\circ} – 40 ^{\circ} ) \times tan 50 ^{\circ} \times tan 80 ^{\circ} ]$

= $\frac{sin 18 ^{\circ} }{ sin 18 ^{\circ} } + \sqrt{ 3 }( \frac{ cot 80 ^{\circ} \times cot 50 ^{\circ} \times tan 50^{\circ}\times tan 80^{\circ}}{ \sqrt{ 3 } } )$

= 1 + $( \frac{ 1 }{tan 80 ^{\circ}} \times \frac{ 1 }{ tan 50 ^{\circ} } \times tan 50^{\circ} \times tan 80 ^{\circ})$ = 1 + 1 = 2 = RHS

(v) $\frac{ 7 \; cos 55 ^{\circ} }{ 3 \; sin 35 ^{\circ}} – \frac{ 4 ( cos 70^{\circ} cosec 20 ^{\circ} ) }{ 3 ( tan 5^{\circ} tan 25^{\circ} tan 45 ^{\circ} tan 65^{\circ}tan 85^{\circ} ) } = 1$

LHS = $\frac{ 7 \; cos 55^{\circ}}{3 \;sin 35^{\circ}}-\frac{ 4 ( cos 70^{\circ}cosec 20^{\circ})}{3(tan 5 ^{\circ} tan 25 ^{\circ}tan 45 ^{\circ} tan 65 ^{\circ}tan 85 ^{\circ} ) }$

= $\frac{ 7 \; cos 55 ^{\circ} } { 3 \; cos ( 90 ^{\circ} – 35 ^{\circ} ) } – \frac{ 4( sin ( 90 ^{\circ} – 70 ^{\circ} ) cosec 20^{\circ} ) } { 3 ( cot ( 90 ^{\circ} – 5 ^{\circ} ) \times cot ( 90 – 25 ^{\circ} ) \times 1 \times tan 65 ^{\circ} \times tan 85 ^{\circ} ) }$

= $\frac{ 7 \; cos 55 ^{\circ} }{ 3 \; cos 55^{\circ} } – \frac{ 4 (sin 20 ^{\circ} \; cosec 20 ^{\circ} ) } {3 ( cot 85 ^{\circ} \; cot 65 ^{\circ} \; tan 65 ^{\circ} \; tan 85 ^{\circ} ) }$

= $\frac{ 7 } { 3 } – \frac{ 4 ( sin 20 ^{\circ} \times \frac{ 1 }{sin 20 ^{\circ} } ) } { 3 (\frac{ 1 }{tan 85 ^{\circ} } \times \frac{ 1 }{ tan 65 ^{\circ} } \times tan 65^{\circ} \times tan 85^{\circ} ) }$

= $\frac{ 7 }{ 3 } – \frac{ 4 }{ 3 }$ = $\frac{ 3 } { 3 }$ = 1 = RHS

Q.5: Prove that:

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

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

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

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

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

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

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

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

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

### RS Aggarwal Class 10 Solutions Chapter 7 – Trigonometric Ratios of Complementary Angles

The Chapter 7 solutions of RS Aggarwal presented here are one of the most important resources for exam preparation. Students who want to prepare the Chapter 7- Trigonometric Ratios of Complementary Angles in a thorough manner can refer to our solutions for the chapter. Using the RS Aggarwal Maths Solution students can find simple to understand explanations for important topics in the chapter. So, just refer to these solutions to prepare well for the exam.

#### Practise This Question

Which of the following is NOT the quality of an ideal source of energy?