Question

Please answer fast. ​

Answer 1

Answer:

$\sqrt{3} \:$

Step-by-step explanation:

solution------->

Given

• perpendicular =1
• hypotenuse =2

plz refer pic for diagram

as per que

ln Δ ABC

by Pythagorus Therom

AC² = AB² + BC²

(2)² = (1)² + BC ²

4 -1 = BC²

taking squre root out we get

BC = √3

now as per given que

$tan (A) = \frac{ac}{ab} \\ \\ tan (A) = \: \frac{1}{ \sqrt{3} }$

$\frac{2 \tan( \ \beta ) }{ 1 - { \tan( \beta ) }^{2} }$

$now \: put \: all \: the \: values \\ \\ \frac{2 \times \frac{1}{ \sqrt{3} } }{1 - { \frac{1}{ \sqrt{3} } }^{2} } \\ \\ \frac{ \frac{2}{ \sqrt{3} } }{ \frac{3 - 1}{3} } \\ \\ \frac{2}{ \sqrt{3} } \times \frac{3}{2} \\ \\ \frac{3}{ \sqrt{3} }$

now

$\frac{3}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \frac{3 \sqrt{3} }{3} \\ \\ 3 \: will \: b \: cancle \\ \\ \sqrt{3}$

Answer 2

Answer:

Option (A) √3 is correct.

Please check the attached image for

Step-by-step explanation: