Question

Prove the following:

`(sqrt(3) + 1) (3 - cot 30^circ)` = tan³ 60° – 2 sin 60°

Answer 1

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.