value
of tan 9/5 find the answer
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Answer:
tan9°−tan27°−tan63°+tan81°
=tan9°+tan81°−(tan27°+tan63°)
=tan9°+tan(90°−9°)−(tan27°+tan(90°−27°))
=tan9°+cot9°−(tan27°+tan27°) [∵cotA=tan(90˚−A)]
=
cos9°
sin9°
+
sin9°
cos9°
−(
cos27°
sin27°
+
sin27°
cos27°
)
=
sin9°cos9°
sin
2
9°+cos
2
9°
−(
sin27°cos27°
sin
2
27°+cos
2
27°
)
=
sin9°cos9°
1
−
sin27°cos27°
1
=
sin18°
2
−
sin54°
2
[∵sin2x=2sinxcosx]
=2[
sin18°sin54°
sin54°−sin18°
]=
sin18°sin(90°−36°)
2.2cos36°sin18°
[∵sinC−sinD=2cos
2
C+D
sin
2
C−D
]
=
cos36°sin18°
2.2cos36°sin18°
=4
Step-by-step explanation:
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