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प्रश्न
Prove the following:
`(sqrt(3) + 1) (3 - cot 30^circ)` = tan3 60° – 2 sin 60°
उत्तर
L.H.S: `(sqrt(3) + 1) (3 - cot 30^circ)`
= `(sqrt3 + 1)(3 - sqrt(3))` ...`[∵ cos 30^circ = sqrt(3)]`
= `(sqrt(3) + 1) sqrt(3) (sqrt(3) - 1)` ...`[∵ (3 - sqrt(3)) = sqrt(3) (sqrt(3) - 1)]`
= `((sqrt(3))^2 - 1) sqrt(3)` ...`[∵ (sqrt(3) + 1)(sqrt(3) - 1) = ((sqrt(3))^2 - 1)]`
= `(3 - 1) sqrt(3)`
= `2sqrt(3)`
Similarly solving R.H.S: tan3 60° – 2 sin 60°
Since, tan 60° = `sqrt(3)` and sin 60° = `sqrt(3)/2`,
We get,
`(sqrt(3))^3 - 2 * (sqrt(3)/2) = 3sqrt(3) - sqrt(3)`
= `2sqrt(3)`
Therefore, L.H.S = R.H.S
Hence, proved.
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