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