Math, asked by sarojagadage, 5 months ago

) tan A+tan (60° + A)+tan (120° + A)=– tan3A​

Answers

Answered by arpitkumar2701
1

Step-by-step explanation:

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

).

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