Get the solution to "X=(1+tan⁡1∘)(1+tan⁡2∘)(1+tan⁡3∘)…(1+tan⁡45∘)tan⁡(90−θ)=cot⁡θ=1tan⁡θ"

Myah Fuller

Answered question

2022-02-27

X = (1 + {\tan 1}^{\circ}) (1 + {\tan 2}^{\circ}) (1 + {\tan 3}^{\circ}) \dots (1 + {\tan 45}^{\circ})

\tan (90 - θ) = \cot θ = \frac{1}{\tan θ}

Chettaf04

Beginner2022-02-28Added 7 answers

Let

θ = 1^{\circ} = \frac{π}{180} rad

. Then

X = \prod_{k = 1}^{45} (1 + \tan (k θ)) = \prod_{k = 1}^{45} \frac{\sin (k θ) + \cos (k θ)}{\cos (k θ)}

= \prod_{k = 1}^{45} \sqrt{2} \frac{\sin ((k + 45) θ)}{\cos (k θ)}

Furthemore

\prod_{k = 1}^{45} \sqrt{2} \frac{\sin ((k + 45) θ)}{\cos (k θ)} = 2^{\frac{45}{2}} \frac{\prod_{k = 1}^{45} \sin ((k + 45) θ)}{\prod_{k = 1}^{45} \cos (k θ)}

= 2^{\frac{45}{2}} \frac{\prod_{n = 0}^{44} \sin (90^{\circ} - n θ)}{\prod_{k = 1}^{45} \cos (k θ)}

Now, using

\sin (90^{\circ} - α) = \cos (α)

2^{\frac{45}{2}} \frac{\prod_{n = 0}^{44} \sin (90^{\circ} - n θ)}{\prod_{k = 1}^{45} \cos (k θ)} = 2^{\frac{45}{2}} \frac{\prod_{n = 0}^{44} \cos (n θ)}{\prod_{k = 1}^{45} \cos (k θ)}

= 2^{\frac{45}{2}} \frac{\cos (0)}{\cos (45^{\circ}}}} = 2^{23}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?