Question

If root3 tan 2theta+ root 3 tan 3theta + tan 2theta tan 3theta = 1, then find the general value of theta
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Answer 1

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Answer 2

Answer:

Ф = ( n + 1 / 6 ) π / 5 where n € Z

Step-by-step explanation:

Given :

√ 3 tan 2 Ф + √ 3 tan 3 Ф + tan 2 Ф. tan 3 Ф = 1

Rewrite as :

√ 3 tan 2 Ф + √ 3 tan 3 Ф = 1 -  tan 2 Ф. tan 3 Ф

Diving by √ 3 both side we get :

√ 3 / √ 3 tan 2 Ф + √ 3 / √ 3 tan 3 Ф = 1 / √ 3 ( 1 -  tan 2 Ф. tan 3 Ф )

tan 2 Ф + tan 3 Ф = 1 / √ 3 ( 1 -  tan 2 Ф. tan 3 Ф )

( tan 2 Ф + tan 3 Ф )  / ( 1 -  tan 2 Ф. tan 3 Ф ) =  1 / √ 3

Now using identity :

tan ( A + B ) = tan A + tan B / 1 - tan A. tan B

tan ( 2 Ф + 3 Ф ) = 1 / √ 3

We know :

tan π / 6 = 1  /√ 3

tan 5 Ф = tan π / 6

On comparing we get :

5 Ф = π / 6

Ф = π / 30

We know for general solution if :

tan Ф = tan α

= > Ф = n π + α where n € Z

So , general solution of given equation is written as :

= > Ф = n π + π / 30

= > Ф = ( n + 1 / 6 ) π / 5 where n € Z

Therefore we get answer.