how can i prove tan9°-tan27°-tan63°+tan81°=4
Answers
Answered by
31
tan9° - tan27° - tan63° + tan81° = 4
LHS = tan9° + tan81° -( tan27° + tan63°)
= {sin9°/cos9° + sin81°/cos81°} - { sin27°/cos27° + sin63°/cos63° }
={ sin9°.cos81°+ sin81°.cos9°}/cos9°.cos81° - { sin27°.cos63° + cos27°.sin63°}/cos27°.cos63°
=sin(9°+ 81°)/cos9°.cos81° - sin(27° +63°)/cos27°.cos63°
= 1/cos9°.cos81° - 1/cos27°.cos63°
=2/(2cos9°.cos81°) - 2/(2cos27°.cos63°)
= 2/{ cos(9° + 81°) + cos(81 -9°) - 2/{ cos(27°+ 63°) + cos(63° -27°)}
=2/cos72° - 2/cos36°
= 2( cos36° - cos72°)/cos36°.cos72°
= 2( 2sin54°.sin18°)/cos36°.cos72°
=4sin54°.sin18°/cos36°.cos72°
=4sin(90-36°).sin(90°-18°)/cos36°.cos72°
=4cos36°.cos72°/cos36°.cos72°
= 4 = RHS
LHS = tan9° + tan81° -( tan27° + tan63°)
= {sin9°/cos9° + sin81°/cos81°} - { sin27°/cos27° + sin63°/cos63° }
={ sin9°.cos81°+ sin81°.cos9°}/cos9°.cos81° - { sin27°.cos63° + cos27°.sin63°}/cos27°.cos63°
=sin(9°+ 81°)/cos9°.cos81° - sin(27° +63°)/cos27°.cos63°
= 1/cos9°.cos81° - 1/cos27°.cos63°
=2/(2cos9°.cos81°) - 2/(2cos27°.cos63°)
= 2/{ cos(9° + 81°) + cos(81 -9°) - 2/{ cos(27°+ 63°) + cos(63° -27°)}
=2/cos72° - 2/cos36°
= 2( cos36° - cos72°)/cos36°.cos72°
= 2( 2sin54°.sin18°)/cos36°.cos72°
=4sin54°.sin18°/cos36°.cos72°
=4sin(90-36°).sin(90°-18°)/cos36°.cos72°
=4cos36°.cos72°/cos36°.cos72°
= 4 = RHS
Aanchal11:
Thank you so much mr passionate
Answered by
6
Answer:
Step-by-step explanation:
Attachments:
Similar questions