# RS Aggarwal Solutions Class 10 Ex 7A

## RS Aggarwal Class 10 Ex 7A Chapter 7

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

#### Practise This Question

What is the nature of the image formed on the retina of a human eye?