Maharashtra State BoardHSC Science (Electronics) 11th
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Prove the following: tanA + tan(60° + A) + tan(120° + A) = 3 tan 3A - Mathematics and Statistics

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Sum

Prove the following:

tanA + tan(60° + A) + tan(120° + A) = 3 tan 3A

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Solution

L.H.S. = tanA + tan(60° + A) + tan(120° + A)

= `tan"A" + (tan60^circ + tan"A")/(1 - tan60^circ*tan"A") + (tan120^circ + tan"A")/(1 - tan120^circ*tan"A")`

= `tan"A" + (sqrt(3) + tan"A")/(1 - sqrt(3)tan"A") + (-sqrt(3) + tan"A")/(1 + sqrt(3)tan"A")    ...[because tan60^circ = sqrt(3) and tan120^circ = tan(180^circ - 60^circ) = -tan60^circ = -sqrt(3)]`

= `(tan"A"(1 - 3tan^2"A") + (sqrt(3) + tan"A")(1 + sqrt(3)tan"A") + (-sqrt(3) + tan"A")(1 - sqrt(3)tan"A"))/((1 - sqrt(3)tan"A")(1 + sqrt(3)tan"A")`

= `(tan"A" - 3tan^3"A" + sqrt(3) + 3tan"A" + tan"A" + sqrt(3)tan^2"A"  - sqrt(3) + 3tan"A" + tan"A" + sqrt(3)tan^2"A")/(1 - 3tan^2"A")`

= `(9tan"A" - 3tan^3"A")/(1 - 3tan^2"A")`

= `3((3tan"A" - tan^3"A")/(1 - 3tan^2"A"))`

= 3 tan 3A

= R.H.S.

Concept: Trigonometric Functions of Sum and Difference of Angles
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 3 Trigonometry - 2
Miscellaneous Exercise 3 | Q II. (14) | Page 57
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