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प्रश्न
Prove the following:
3tan610° – 27 tan410° + 33tan210° = 1
उत्तर
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
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