# RS Aggarwal Class 10 Solutions Chapter 8 Trigonometric Identities

## RS Aggarwal Class 10 Chapter 8 Trigonometric Identities Solutions Free PDF

The trigonometric identities are simply the identities that helps establish a relationship between different trigonometric functions. Learning these identities will undoubtedly prove useful for all the students, who are preparing for this particular chapter of Trigonometric Identities. The inverse of a few identities is equal to sum of the trigonometric identities. Check out the various types of identities below:

1. Cos(x)= 1/Sec(x)
2. Sec(x)= 1/Cos(x)
3. Sin(x)= 1/Cosec(x)
4. Cosec(x)= 1/Sin(x)
5. Cot(x)= 1/Tan(x)

Check out the RS Aggarwal Class 10 Solutions Chapter 8 Trigonometric Identities below:

Q.1: Without using trigonometric table, evaluate:

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

Sol:

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

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

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

Sol:

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

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

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

Sol:

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

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

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

Sol:

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

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

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

Sol:

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

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

Sol:

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

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

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

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

Sol:

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

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

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

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

Sol:

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

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

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

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

Sol:

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

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

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

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

Sol:

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

Sol:

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

Sol:

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

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

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

Sol:

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

Sol:

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

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

Sol:

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

Sol:

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

Sol:

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

Sol:

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

Sol:

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

Sol:

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

(i) 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 â€“ 2 * $cos 70 ^{\circ} \times \frac{ 1 }{ cos 70 ^{\circ} }$

= 2 â€“ 2 = 0 = RHS

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

Sol:

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

Sol:

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

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

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

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

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

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

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

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

Sol:

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

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

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

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

= $\frac{ 7 }{ 3 } – \frac{ 4 }{ 3 }$

= $\frac{ 3 } { 3 }$ = 1 = RHS

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