 RS Aggarwal Class 10 Solutions Chapter 7 - Trigonometric Ratios Of Complementary Angles Ex 7A(7.1)

RS Aggarwal Class 10 Chapter 7 - Trigonometric Ratios Of Complementary Angles Ex 7A(7.1) Solutions Free PDF

The Chapter 7 solutions of RS Aggarwal for Class 10 are made available for students with the opportunity to prepare the chapter thoroughly. All the solutions mentioned below for Chapter 7 have been developed according to the latest syllabus of Class 10 Maths suggested by the CBSE. Students should value these solutions as it will help them to clear their doubts and develop their question-solving ability.

For better understanding, all the RS Aggarwal Class 10 solutions for Chapter 7 are given in a simple and structured manner. It will help you to develop a thorough understanding of each topic. So, if you want to prepare well for the exam, then referring to these solutions will guide you in the best direction.

Download PDF of RS Aggarwal Class 10 Solutions Chapter 7– Trigonometric Ratios of Complementary Angles Ex 7A (7.1)

Q.1: Prove that:

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

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

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

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

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

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

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.2: Prove that:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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: Is $sec 2A = cosec ( A – 42^{\circ} )$ , 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

Key Features of RS Aggarwal Class 10 Solutions Chapter 7 – Trigonometric Ratios Of Complementary Angles Ex 7A (7.1)

• It helps in improving their overall academic performance.
• They can refer to these solutions if they are stuck or have any doubt while solving the questions.
• The solutions are explained in simple language with options to solve a particular type of questions.
• Students are advised to refer RS Aggarwal Maths solutions while preparing for their final exam.

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