Maharashtra State BoardHSC Science (General) 11th
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Prove the following: 3tan610° – 27 tan410° + 33tan210° = 1 - Mathematics and Statistics

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Sum

Prove the following:

3tan610° – 27 tan410° + 33tan210° = 1

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Solution

Since, tan30° = `1/sqrt(3)`

∴ tan3(10°) = `1/sqrt(3)`

∴ `(3tan10^circ - tan^3 10^circ)/(1 - 3tan^2 10^circ) = 1/sqrt(3)`

Squaring both the sides, we get

`((3tan10^circ - tan^3 10^circ)^2)/((1 - 3tan^2  10^circ)^2) = 1/3`

∴ `(9tan^2 10^circ - 6tan^4 10^circ + tan^6 10^circ)/(1 - 6tan^2 10^circ + 9tan^4 10^circ) = 1/3`

∴ 3(9tan210° – 6tan410°+ tan610°) = 1 – 6tan210° + 9tan410°

∴ 27tan210° – 18tan410° + 3tan610° = 1 – 6tan210° + 9tan410°

∴ 3tan610° – 27tan410° + 33tan210° = 1

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. (15) | Page 57
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