Question

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

Answer 1

SOLUTION :  

The correct option is (a) : sin 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 / 4/3

= (2/√3) × (¾)

= 3/2√3

= (3 × √3) / (2√3  × √3)

[By rationalising the denominator]

= 3√3 / 2×3

= √3/2

= sin 60°           [sin 60° = √3/2]

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

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

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

$= > \frac{2tan30^{0}}{1+tan^{2}30^{0}} \\ \\ (tan60^{0} = \frac{1}{ { \sqrt{3} } }) \\ \\ = > \frac{2 \times \frac{1}{ { \sqrt{3} } } }{1 + { (\frac{1}{ \sqrt{3} } )}^{2} } \\ \\ = > \frac{ \frac{2}{ \sqrt{3} } }{1 +\frac{1}{3} } \\ \\ = > \frac{ \frac{2}{ \sqrt{3} } }{ \frac{3 + 1}{3} } \\ \\ = > \frac{ \frac{2}{ \sqrt{3} } }{ \frac{4}{3} } \\ \\ = > \frac{2}{ \sqrt{3} } \times \frac{3}{4} \\ \\ = > \frac{\sqrt{3}}{2}\\ \\ = > sin60^{0} \\ (sin60^{0} = \frac{\sqrt{3}}{2})$

$\boxed{Answer \: : \: Option(a) \: sin60^{0} }$

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