Proved that, tan(A+60°)+ tan(A-60°)=4sin2A/(1-4sin^2A)
Answers
Answered by
0
Answer:
ANSWER
tan(60+A)=
1−tan60
o
tanA
tan60
o
+tanA
=
1−
3
tanA
3
+tanA
tan(60−A)=
1+tan60
o
tanA
tan60
o
−tanA
=
1+
3
tanA
3
−tanA
tan(60+A)−tan(60−A)=
1−
3
tanA
3
+tanA
−
1+
3
tanA
3
−tanA
=
1
2
−(
3
tanA)
2
(
3
+tanA)(1+
3
tanA)−(
3
−tanA)(1−
3
tanA)
=
1−3tan
2
A
3
+3tanA+tanA+
3
tan
2
A−
3
+3tanA+tanA−
3
tan
2
A
=
1−3tan
2
A
8tanA
tanA+tan(60+A)−tan(60−A)=tanA+
1−3tan
2
A
8tanA
=
1−3tan
2
A
tanA−3tan
3
A+8tanA
=
1−3tan
2
A
3(3tanA−tan
3
A)
=3tan3A
(∵tan3A=
1−3tan
2
A
3tanA−tan
3
A
).
see the image will be helpful for you
Attachments:
Answered by
0
Answer:
I hope answer is helpful to us
Attachments:
Similar questions