2 - Insightsfunda

Trigonometry Practice Set-4
TRIGONOMETRIC IDENTITIES
Prove the following identities:
(1)
(1 − sin 2 A ) sec2 A = 1
(6)
(1 + tan A ) cos A = 1
( cosec A − 1) tan A = 1
(1 − cos A ) cosec A = 1
(1 + cot A ) sin A = 1
( sec A − 1) cot A = 1
(7)
cot 2 θ −
(2)
(3)
(4)
(5)
2
2
2
2
2
2
2
2
2
2
(8)
1
= −1
sin 2 θ
sec A 1 − sin 2 A = 1
(9)
(10)
cosec A 1 − cos 2 A = 1
(1 − cos A )(1 + cos A ) (1 + cot 2 A ) = 1
(11)
(12)
(13)
(1 − cos A ) sec A = tan A
( sec A − 1)( cosec A − 1) = 1
2
2
2
2
2
sec A (1 − sin A )( sec A + tan A ) = 1
(14) (1 + tan 2 θ ) (1 − sin θ )(1 + sin θ )
(15)
sec A (1 + sin A )( sin A − tan A ) = 1
2cos 2 − 1
(16) cos θ − tan θ =
sin θ cos θ
2
1 − tan θ
(17)
= tan 2 θ where θ ≠ 45°
2
cos θ − 1
(18)
sec2 θ + cosec2θ = ( tan θ + cot θ )
(19)
cos θ
1 − sin θ
=
1 + sin θ
cos θ
(24)
1
= sec θ + tan θ
sec θ − tan θ
sec θ − 1 1 − cos θ
=
sec θ + 1 1 + cos θ
1 − sin θ
2
= ( sec θ − tan )
1 + sin θ
sec θ − tan θ
= 1 − 2sec θ tan + 2tan 2 θ
sec θ + tan θ
2
( sin θ − cos θ ) = (1 − 2sin θ cos θ )
(25)
(26)
(27)
(28)
sec 2 θ + cosec 2θ = sec 2 θ cosec2θ
tan 2 θ + cot 2 θ + 2 = sec 2 θ cosec 2θ
tan 2 θ − sin 2 θ = tan 2 θ ⋅ sin 2 θ
(1 + cot θ − cosec θ )(1 + tan θ + sec θ ) = 2
(20)
(21)
(22)
(23)
(29)
( cosec θ − sin θ )( sec θ − cos θ )( tan θ + cot θ ) = 1
cosec A
cosec A
+
= 2sec 2 A
cosec A − 1 cosec A + 1
tan A
cot A
(31)
+
= sec A cosecA + 1
1 − cot A 1 − tan A
2
2
 1 + sin 2 θ 
1  
1 

(32)  tan θ +
tan
θ
−
=
2


2
cos θ  
cos θ 

 1 − sin θ 
1
1
1
1
(33)
−
=
−
( cosecθ + cot θ ) sin θ sin θ ( cosecθ − cot θ )
1
1
1
1
(34)
−
=
−
( sec θ + tan θ ) cos θ cos θ ( sec θ − tan θ )
tan A + sec A − 1 1 + sin A
(35)
=
tan A − sec A + 1
cos A
1
(36) ( cosecθ − sin θ )( sec θ − cos θ ) =
tan θ + cot θ
1 
1 
1

(37)  1 +
(38)
 1 +
=
2
2
2
tan θ 
cot θ  sin θ − sin 4 θ

sin A − sinB cos A − cosB
=
=0
cos A + cosB sin A + sin B
(39) sin 8 θ − cos 8 θ = ( sin 2 θ − cos 2 θ )(1 − 2sin 2 θ cos 2 θ )
(30)
(40)
(41)
(42)
(44)
(45)
(47)
(48)
(49)
(50)
(51)
(52)
(53)
(54)
(55)
(56)
(57)
(58)
(59)
(60)
sin 2 A − sin 2 B
tan A tan B =
cos 2 A ⋅ cos 2 B
2
2
(1 + tan A tan B ) + ( tan A − tan B ) = sec2 Asec2 B
2
2
(43)
sec6 θ = tan 6 θ + 3tan 2 θ sec 2 θ + 1
2
2
2
2
tan A + cot A = sec A ⋅ cosec A − 2
( sec A − cosec A )(1 + tan A + cot A ) = tan Asec A − cot A cosecA
tan 2 Asec 2 B − sec 2 A tan 2 B = tan 2 A − tan 2 B
2
 1 + sin θ − cos θ  1 − cos θ

 =
1
+
sin
θ
+
cos
θ

 1 + cos θ
(46)
( tan A + cosec B ) − ( cot B − sec A ) = 2tan A cot B ( cosec A + secB )
2
2
2
( sin A + sec A ) ( cos A + cosec A ) = (1 + sec A ⋅ cosec A )
(1 + cot A + tan A )( sin A − cos A ) = sin 2 A ⋅ cos 2 A
2
2
sec 3 A − cosec 3 A
1 + cos θ + sin θ 1 + sin θ
=
1 + cos θ − sin θ
cos θ
2
1 + cos θ − sin θ
= cot θ
sin θ (1 + cos θ )
1
1
1 − sin 2 θ ⋅ cos 2 θ

 2
2
+
 2
 sin θ ⋅ cos θ =
2
cosec 2θ − sin 2 θ 
2 + sin 2 θ ⋅ cos 2 θ
 sec θ − cos
sin A
cos A
+
=1
sec A + tan A − 1 cosecA + cot A − 1
1 + sin θ cot θ + cos θ
=
1 − sin θ cot θ − cos θ
sin 4 θ − cos 4 θ = sin 2 θ − cos 2 θ = 2sin 2 θ − 1 = 1 − 2cos 2 θ
(
) (
) (
( sin θ + cos θ ) = (1 − 2sin θ cos θ )
( sec θ − sec θ ) = ( tan θ + tan θ )
( sin θ + cos θ ) = (1 − 3sin θ cos θ )
4
4
4
2
6
6
2
4
2
2
2
2
cos A
sin 2 A
−
= sin Acos A
1 − tan A cos A − sin A
2 ( sin 6 θ + cos 6 θ ) − 3 ( sin 4 θ + cos 4 θ ) + 1 = 0
) (
)
(61)
(62)
(63)
(64)
(65)
(66)
(67)
(68)
cos 3 θ + sin 3 θ cos 3 θ − sin 3 θ
+
=2
cos θ + sin θ
cos θ − sin θ
sin θ − 2sin 3 θ
= tan θ
2cos 3 θ − cos θ
tan A + tan B
= tan A tan B
cot A + cot B
sin θ + cos θ sin θ − cos θ
2
+
=
2
sin θ − cos θ sin θ + cos θ ( sin θ − cos 2 θ )
1 + sin A
1
=
+ tan A
1 − sin A cos A
tan 3 θ
cot 3
+
= sec θ cosecθ − 2sin θ cos θ
1 + tan 2 θ 1 + cot 2 θ
sin 2 Acos 2 B + cos 2 Asin 2 B + cos 2 Acos 2 B + sin 2 Asin 2 B = 1
sin 2 Acos 2 B − cos 2 Asin 2 B = sin 2 A − sin 2 B