Simplifying trigonometry function by substitution

Question

I want to simplify the following expression: $\frac{{\left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + {{\tan }^2}{1^{\circ}}} \right)\left( {1 + {{\tan }^2}{2^{\circ}}} \right)}}$

My approach is as follow

$T = \left( {1 + \sqrt 3 \tan {1^{\circ}}} \right)\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right)\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$

$\Rightarrow \tan {58^{\circ}}\left( {1 + \sqrt 3 \tan {2^{\circ}}} \right) = \sqrt 3 - \tan {2^{\circ}}$

$\Rightarrow \tan {59^{\circ}}\left( {1 + \sqrt 3 \tan {2^0}} \right) = \sqrt 3 - \tan {1}$

$\tan {60^{\circ}} = \frac{{\left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right)}}{{\left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)}} \Rightarrow \left( {\tan {1^{\circ}} + \tan {{59}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right)$

$\tan {60^{\circ}} = \frac{{\left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right)}}{{\left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)}} \Rightarrow \left( {\tan {2^{\circ}} + \tan {{58}^{\circ}}} \right) = \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right)$

$T = \frac{{\sqrt 3 - \tan {1^{\circ}}}}{{\tan {{59}^{\circ}}}} \times \frac{{\sqrt 3 - \tan {2^{\circ}}}}{{\tan {{58}^{\circ}}}} \times \sqrt 3 \left( {1 + \tan {{59}^{\circ}}\tan {1^{\circ}}} \right) \times \sqrt 3 \left( {1 + \tan {{58}^{\circ}}\tan {2^{\circ}}} \right){\cos ^2}{1^{\circ}}{\cos ^2}{2^{\circ}}$

After this step I am confused

@SamarImamZaidi you can use ^\circ instead of ^0 (which may be confusing). — an4s, Jul 21 '20 at 16:16
Sir, I copied the latex option through MathType so I just wrote the formula and then it converted it into Latex — Samar Imam Zaidi, Jul 21 '20 at 16:18
This question is Integer type hence at the end we will get an Inegral Value after substitution — Samar Imam Zaidi, Jul 21 '20 at 16:19
Related https://math.stackexchange.com/questions/188746/calculating-sqrt3-tan-1-circ-sqrt3-tan2-circ-sqrt — lab bhattacharjee, Jul 21 '20 at 18:24
@lab bhattacharjee Can you approach the problem from where i left. — Samar Imam Zaidi, Jul 21 '20 at 20:24

score 1 · Accepted Answer · edited Jul 21 '20 at 16:22

1

$$ \tan 59^{\circ} = \frac{\sqrt{3}-\tan 1^{\circ}}{1+\sqrt{3} \tan 1^{\circ}} $$ $$ \tan 59^{\circ}+\tan 1^{\circ} = \frac{\sqrt{3}(1+\tan^2 1^{\circ})}{1+\sqrt{3} \tan 1^{\circ}}, $$so we have $$ (1+\sqrt{3} \tan 1^{\circ})(\tan 59^{\circ}+\tan 1^{\circ})=\sqrt{3}(1+\tan^2 1^{\circ}) $$A similar argument works for $\tan 2^{\circ}$ and $\tan 58^{\circ}$. So the answer is $3$.

edited Jul 21 '20 at 16:22

Integrand

8,457

answered Jul 21 '20 at 16:19

Namburu Karthik

329

Thanks for the edit – Namburu Karthik Jul 21 '20 at 16:22
2

Type it down and learn mathjax, i see you have been around for quiet a while now. – Arjun Jul 21 '20 at 16:24
Yeah Integrand edited it and helped me – Namburu Karthik Jul 21 '20 at 16:25
3

Do it yourself, no one is bound to do it for you, you have a lot of answers most of which are poorly formatted – Arjun Jul 21 '20 at 16:28
Anyway the answer is calculated but i would like soneone to approach fron where i left – Samar Imam Zaidi Jul 21 '20 at 19:31

score 0 · Answer 2 · answered Jul 22 '20 at 15:41

0

Generalization:

$$\dfrac{(1+\tan y\tan z)(\tan z+\tan(y-z))}{1+\tan^2z} =\cdots =\dfrac{\cos(y-z)\sin(z+y-z)\cos^2z}{\cos y\cos z\cos z\cos(y-z)} =\tan y$$

$$\text{ if }\cos(y-z)\cos z\ne0$$

Here $y=60^\circ$

Then $z=1^\circ, 2^\circ$

answered Jul 22 '20 at 15:41

lab bhattacharjee

274,582

Simplifying trigonometry function by substitution

2 Answers2