My situation is as follows. Can the expression from below be simplified using precalculus and plain by hand calculation without requiring a calculator?

$$B=\sqrt{3} \tan 70^{\circ}- 4 \sin 70^{\circ}+1$$

What I attempted to do was to split the functions in a sum of $30^{\circ}+40^{\circ}$ since the trigonometric expressions for $30^{\circ}$ is 'known'.

By going into that route I went through this shown below:

$\sqrt{3} \tan\left(30+40\right)-4\sin\left(30+40\right)+1$

$\sqrt{3}\left(\frac{\tan(30)+\tan(40)}{1-\tan(30)\tan(40)}\right)-4(\sin(30)\cos(40)+\cos(30)\sin(40)+1$

$\sqrt{3}\left(\frac{\frac{1}{\sqrt 3}+\tan(40)}{1-\frac{1}{\sqrt 3}\tan(40)}\right)-4\left(\frac{1}{2}\cos(40)+\frac{\sqrt 3}{2}\sin(40)\right)+1$

$\sqrt{3}\left(\frac{1+\sqrt 3\tan(40)}{\sqrt 3-\tan(40)}\right)-2\cos(40)-2\sqrt {3} \sin(40)+1$

$\frac{\sqrt 3 + 3\frac{\sin(40)}{\cos(40)}}{\sqrt 3-\frac{\sin(40)}{\cos(40)}}-2\cos(40)-2\sqrt {3} \sin(40)+1$

$\frac{\sqrt 3 \cos (40) + 3 \sin(40)}{\sqrt 3 \cos (40)-\sin(40)}-2\cos(40)-2\sqrt {3} \sin(40)+1$

Then multiplying by $\sqrt 3 \cos (40)-\sin(40)$

$\frac{\sqrt 3 \cos (40) + 3 \sin(40)-2\sqrt 3\cos^2(40)+2\sin(40)\cos(40)-6\sin(40)\cos(40)+2\sqrt{3}\sin^2(40)+\sqrt 3 \cos (40)-\sin(40)}{\sqrt 3 \cos (40)-\sin(40)}$

$\frac{2\sqrt 3 \cos (40) + 2 \sin(40)-2\sqrt 3\cos(80)-2\sin(80)}{\sqrt 3 \cos (40)-\sin(40)}$

Now dividing by $4$ on the numerator:

$\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{4\sqrt 3 \cos (40)-4\sin(40)}$

$\frac{\sin(100)-\sin(140)}{4\sqrt 3 \cos (40)-4\sin(40)}$

Then dividing by $8$ in the denominator

$\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\frac{\sqrt 3}{2} \cos (40)-\frac{1}{2}\sin(40)}$

$\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\sin(60) \cos (40)-\cos(60)\sin(40)}$

$\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\sin(20)}$

Finally using prosthaphaeresis identities:

$\frac{\frac{1}{8}\left(2\cos(60)\sin(20)\right)}{\sin(20)}$

So I end up with:

$\frac{1}{8}\left(2\cos(60^{\circ})\right)=\frac{1}{4}\left(\frac{1}{2}\right)=\frac{1}{8}$

But I'm not sure whether this is an adequate method either. Does there exist a better way to simplify it or to ease calculations? Can somebody help me with an easier and quicker procedure?

Apparently, the answer according to a calculator is 2. But I'm like playing Where's Wally with where I made a mistake. I could catch up that when I divided by 4 in the denominator there was a mistake, so by continuing from there:

$\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{\frac{1}{4}\left(\sqrt 3 \cos (40)-\sin(40)\right)}$.

By continuing the computation by hand, I was able to simplify the expression to:

\(\displaystyle \frac{\sin 100 - \sin 140}{\frac{1}{2}\left(\sin(20)\right)}=1\)

But for some reason I can't find where it is the mistake.

$$B=\sqrt{3} \tan 70^{\circ}- 4 \sin 70^{\circ}+1$$

What I attempted to do was to split the functions in a sum of $30^{\circ}+40^{\circ}$ since the trigonometric expressions for $30^{\circ}$ is 'known'.

By going into that route I went through this shown below:

$\sqrt{3} \tan\left(30+40\right)-4\sin\left(30+40\right)+1$

$\sqrt{3}\left(\frac{\tan(30)+\tan(40)}{1-\tan(30)\tan(40)}\right)-4(\sin(30)\cos(40)+\cos(30)\sin(40)+1$

$\sqrt{3}\left(\frac{\frac{1}{\sqrt 3}+\tan(40)}{1-\frac{1}{\sqrt 3}\tan(40)}\right)-4\left(\frac{1}{2}\cos(40)+\frac{\sqrt 3}{2}\sin(40)\right)+1$

$\sqrt{3}\left(\frac{1+\sqrt 3\tan(40)}{\sqrt 3-\tan(40)}\right)-2\cos(40)-2\sqrt {3} \sin(40)+1$

$\frac{\sqrt 3 + 3\frac{\sin(40)}{\cos(40)}}{\sqrt 3-\frac{\sin(40)}{\cos(40)}}-2\cos(40)-2\sqrt {3} \sin(40)+1$

$\frac{\sqrt 3 \cos (40) + 3 \sin(40)}{\sqrt 3 \cos (40)-\sin(40)}-2\cos(40)-2\sqrt {3} \sin(40)+1$

Then multiplying by $\sqrt 3 \cos (40)-\sin(40)$

$\frac{\sqrt 3 \cos (40) + 3 \sin(40)-2\sqrt 3\cos^2(40)+2\sin(40)\cos(40)-6\sin(40)\cos(40)+2\sqrt{3}\sin^2(40)+\sqrt 3 \cos (40)-\sin(40)}{\sqrt 3 \cos (40)-\sin(40)}$

$\frac{2\sqrt 3 \cos (40) + 2 \sin(40)-2\sqrt 3\cos(80)-2\sin(80)}{\sqrt 3 \cos (40)-\sin(40)}$

Now dividing by $4$ on the numerator:

$\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{4\sqrt 3 \cos (40)-4\sin(40)}$

$\frac{\sin(100)-\sin(140)}{4\sqrt 3 \cos (40)-4\sin(40)}$

Then dividing by $8$ in the denominator

$\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\frac{\sqrt 3}{2} \cos (40)-\frac{1}{2}\sin(40)}$

$\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\sin(60) \cos (40)-\cos(60)\sin(40)}$

$\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\sin(20)}$

Finally using prosthaphaeresis identities:

$\frac{\frac{1}{8}\left(2\cos(60)\sin(20)\right)}{\sin(20)}$

So I end up with:

$\frac{1}{8}\left(2\cos(60^{\circ})\right)=\frac{1}{4}\left(\frac{1}{2}\right)=\frac{1}{8}$

But I'm not sure whether this is an adequate method either. Does there exist a better way to simplify it or to ease calculations? Can somebody help me with an easier and quicker procedure?

Apparently, the answer according to a calculator is 2. But I'm like playing Where's Wally with where I made a mistake. I could catch up that when I divided by 4 in the denominator there was a mistake, so by continuing from there:

$\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{\frac{1}{4}\left(\sqrt 3 \cos (40)-\sin(40)\right)}$.

By continuing the computation by hand, I was able to simplify the expression to:

\(\displaystyle \frac{\sin 100 - \sin 140}{\frac{1}{2}\left(\sin(20)\right)}=1\)

But for some reason I can't find where it is the mistake.

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