RS Aggarwal Solutions Class 10 Ex 8A

All these RS Aggarwal class 10 solutions Chapter 8 Exercise 8.1 : Trigonometric Identities are solved by Byju's top ranked professors as per CBSE guidelines.

Q.1: Prove that:

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

Sol:

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

Sol:

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

Hence Proved

 

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

Sol:

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

Sol:

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

Sol:

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

Sol:

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

Sol:

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

??

Q.2: Prove that:

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

Sol:

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

Sol:

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

??

Sol:

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

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

Sol:

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

Sol:

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.3: Prove that:

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

Sol:

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

Sol:

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

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

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

 

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

Sol:

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

Sol:

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

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

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

 

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

Sol:

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

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

Sol:

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

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

= \(cos 23 ^{\circ} + sin 15 ^{\circ}\)

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

Sol:

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

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

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

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

Sol:

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

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

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

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

Sol:

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

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

= \( sec 36^{ \circ} \; ??+ ??\; cos18^{ \circ} \)

??

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

Hence Proved

 


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