NCERT Solutions
Class 10 Maths
Introduction to Trigonometry
Ex.8.3 Q.4
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(1) (cosec θ – cot θ)2 = 
(2) 
+ 
= 2 sec A
(3) 
+ 
= 1 + sec θ + cosec θ
[Hint: Write the expression in terms of sin θ and cos θ]
(4) 
= sin2 A ÷ (1 – cos A)
[Hint: Simplify LHS and RHS separately]
(5) (cos A – sin A + 1) ÷ (cos A + sin A - 1) = cosec A, using the identity cosec2 A = 1 + cot2 A.
(6) √ {(1 + sin A) ÷ (1 – sin A)} = sec A + tan A
(7) (sin θ – 2 sin3 θ ÷ (2cos3 θ – cos θ) = tan θ
(8) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
(9) (cosec A – sin A) (sec A – cos A) = 
[Hint: Simplify LHS and RHS separately]
(10) (1 + tan2 A) ÷ (1 + cot2 A) = (1 - tan2 A) ÷ (1 - cot2 A) = tan2 A
(1) LHS:
(Cosec θ – cot θ)2
= ( 
- 
)2
= {(1 – cos θ) ÷ sin θ}2
= (1 – cos θ)2 ÷ sin2 θ
= (1 – cos θ)2 ÷ (1 – cos2 θ)
[sin2 θ + cos2 θ = 1]
= [(1 – cos θ)2] ÷ [{(1 – cos θ) (1 + cos θ)}]
[a2 – b2 = (a - b) (a + b)]
= [(1 – cos θ) ÷ {(1 + cos θ)]
= RHS
(2) LHS:
+ 
= {cos2 A + (1 + sin A)2} ÷ {(1 + sin A) × cos A}
= {cos2 A + 1 + sin2 A + 2sin A} ÷ {(1 + sin A) × cos A}
= {1 + 1 + 2sin A} ÷ {(1 + sin A) × cos A}
[sin2 θ + cos2 θ = 1]
= {2 + 2sin A} ÷ {(1 + sin A) × cos A}
= 2(1 + sin A) ÷ {(1 + sin A) × cos A}
= 
= 2 sec A
= RHS
(3) LHS:
+ 
= [(
) ÷ (1 – 
)] + [( )÷(1 – )]
= ( ) ÷ { 
} + ( )÷{ 
}
= [sin2 θ ÷ {cos θ (sin θ – cos θ)}] + [cos2 θ ÷ {sin θ (cos θ – sin θ)}]
= [sin2 θ ÷ {cos θ (sin θ – cos θ)}] – [cos2 θ ÷ {sin θ (sin θ – cos θ)}]
= (sin3 θ - cos3 θ) ÷ {cos θ × sin θ × (sin θ – cos θ)}
= {(sin2 θ + cos2 θ + sin θ cos θ) (sin θ – cos θ)} ÷ {cos θ × sin θ × (sin θ – cos θ)}
= (1 + sin θ cos θ) ÷ (cos θ × sin θ)
[sin2 θ + cos2 θ = 1]
= [1 ÷ (cos θ × sin θ)] + [(cos θ × sin θ) ÷ (cos θ × sin θ)]
= sec θ cosec θ + 1
= RHS
(4) LHS
= (1 + 
) ÷ ( 
)
= {
} ÷ (
)
= (1 + cos A) ÷ 1
= [(1 + cos A) ÷ 1] × [
]
= (1 – cos2 A) ÷ (1 + cos A)
= 
[sin2 θ + cos2 θ = 1]
= RHS
(5) LHS:
= {
}÷{ 
}
[divide by sin A]
= (cot A – 1 + cosec A) ÷ (cot A + 1 - cosec A)
= {cot A + (-1) + cosec A} ÷ (cot A + 1 - cosec A)
= {cot A + cosec A + cosec2 A - cot2 A} ÷ (cot A + 1 - cosec A)
= {cot A + cosec A + (cosec A - cot A) (cosec A + cot A)} ÷ (cot A + 1 - cosec A)
= {(cosec A + cot A) (1 - cosec A + cot A)} ÷ (cot A + 1 - cosec A)
= cosec A + cot A
= RHS
(6) LHS:
√ {(1 + sin A) ÷ (1 – sin A)} = √ [{(1 + sin A) ÷ (1 – sin A)} × {(1 + sin A) ÷ (1 + sin A)}]
= √ {(1 + sin A)2÷ (1 – sin2 A)}
= √ {(1 + sin A)2 ÷ cos2 A}
=
= 
+ 
= sec A + tan A = RHS
(7) (sin θ – 2 sin3 θ) ÷ (2cos3 θ – cos θ)
= {sin θ (1 – 2 sin2 θ)} ÷ {cos θ (2cos2 θ – 1)}
= {sin θ (1 – 2 sin2 θ)} ÷ {cos θ × 2(1 - sin2 θ) – 1)}
= {sin θ (1 – 2 sin2 θ)} ÷ {cos θ (2 - 2sin2 θ – 1)}
= {sin θ (1 – 2 sin2 θ)} ÷ {cos θ (1 - 2sin2 θ)}
= 
= tan θ
= RHS
(8) LHS:
(sin A + cosec A)2 + (cos A + sec A)2
= sin2 A + cosec2 A + 2sin A cosec A + cos2 A + sec2 A + 2cos A cot A
= (sin2 A + cos2 A) + (cosec2 A + sec2 A) + 2sin A × 
+ 2cos A × 
= 1 + 1 + cot2 A + 1 + tan2 A + 2 + 2
= 7 + cot2 A + tan2 A
= RHS
(9) LHS:
(cosec A – sin A) (sec A – cos A) = ( 
– sin A) ( – cos A)
= {(1 – sin2 A) ÷ sin A} {(1 – cos2 A) ÷ cos A}
= {cos2 A) ÷ sin A} {sin2 A) ÷ cos A}
= sin A cos A …………... (1)
RHS:
= 1 ÷ (
+ 
)
= 1 ÷ {(sin2 A + cos2 A) ÷ (sin A cos A)}
= 1 ÷ {1 ÷ (sin A cos A)}
= sin A cos A …………… (2)
From equation (1) and (2), we get
(cosec A – sin A) (sec A – cos A) =
(10) LHS:
(1 + tan2 A) ÷ (1 + cot2 A) = sec2 A ÷ cosec2 A
= (1 ÷ cos2 A) ÷ (1 ÷ sin2 A)
= sin2 A ÷ cos2 A
= tan2 A
= RHS
Again (1 - tan A) 2 ÷ (1 - cot A) 2 = {(1 - tan A) ÷ (1 - cot A)}2
= {(1 -
) ÷ (1 - 
)} 2
= [{(cos A - sin A) ÷ cos A} ÷ {(sin A - cos A) ÷ sin A)}] 2
= [{(cos A - sin A) ÷ cos A} × {sin A ÷
}] 2
= [- {(sin A - cos A) ÷ cos A} × {sin A ÷ 
}] 2
= ( )2
= 
= tan2 A
= RHS