Question

$\frac{2tan30^{0}}{1-tan^{2}30^{0}}$ is equal to
(a)cos 60°
(b)sin 60°
(c)tan 60°
(d)sin 30°

Answer 1

SOLUTION :  

The correct option is (c) : tan 60°

Given : 2 tan 30° / 1 -  tan² 30°  

2 tan 30° / 1 - tan² 30°  

= 2 (1/√3) / 1 -  (1/√3)²

[tan 30° = 1/√3 ]

= 2/√3 /  1 - ⅓

= 2/√3 / (3 - 1)/3

= 2/√3 / 2/3

= (2/√3) × (3/2)

= 3/√3

= (3 × √3) / (√3 × √3)

[By rationalising the denominator]

= 3√3 / 3  

= √3

= tan 60°           [tan 60° = √3]

Hence, the value of 2 tan 30° / 1 - tan² 30° is tan 60° .  

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\underline{\large\bf{\mathfrak{Hello!}}}$

$= > \frac{2tan30^{0}}{1-tan^{2}30^{0}} \\ \\ (tan60^{0} = \frac{1}{ { \sqrt{3} } }) \\ \\ = > \frac{2 \times \frac{1}{ { \sqrt{3} } } }{1 - { (\frac{1}{ \sqrt{3} } )}^{2} } \\ \\ = > \frac{ \frac{2}{ \sqrt{3} } }{1 - \frac{1}{3} } \\ \\ = > \frac{ \frac{2}{ \sqrt{3} } }{ \frac{3 - 1}{3} } \\ \\ = > \frac{ \frac{2}{ \sqrt{3} } }{ \frac{2}{3} } \\ \\ = > \frac{2}{ \sqrt{3} } \times \frac{3}{2} \\ \\ = > \sqrt{3} \\ \\ = > tan60^{0} \\ (tan60^{0} = \sqrt{3})$

$\boxed{Answer \: : \: Option(c) \: tan60^{0} }$

$\bf{\mathfrak{Hope \: this \: helps...:)}}$