RS Aggarwal Class 10 Solutions Trigonometric Ratios of Complementary Angles

RS Aggarwal Class 10 Solutions Chapter 7

All these RS Aggarwal class 10 solutions Chapter 7 Trigonometric Ratios of Complementary Angles are solved by Byju's top ranked professors as per CBSE guidelines.

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