4 years ago

Which point best approximates √3? A B C D

Mathematics

1 answer:

noname [10]4 years ago

7 0

Answer:

Step-by-step explanation:

Square root of 3 is between 1 and 2

1^2 =1

(sqrt(3)) ^2 =3

2^2 = 4

So it must be point B

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Question is attached.<br>Step by step explanation only!

il63 [147K]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{\tan\theta}{ 1- \cot \theta} + \frac{\cot\theta}{1 - \tan\theta} = \frac{\dfrac{\sin\theta}{\cos\theta}}{1 - \dfrac{\cos\theta}{\sin\theta}} + \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}}$

Recall that tanθ = sinθ / cosθ. Likewise, cotθ = cosθ / sinθ. Then by substitution:

$\displaystyle \begin{aligned} \frac{\tan\theta}{ 1- \cot \theta} + \frac{\cot\theta}{1 - \tan\theta} & = \frac{\dfrac{\sin\theta}{\cos\theta}}{1 - \dfrac{\cos\theta}{\sin\theta}} + \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}} \\ \\ & \stackrel{\checkmark}{=} \frac{\dfrac{\sin\theta}{\cos\theta}}{1 - \dfrac{\cos\theta}{\sin\theta}} + \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}} \end{aligned}$