Expert answers to "If sin⁡x+cos⁡x=3+12 then tan⁡x+cot⁡x= ?"

Tomasz Gibson

Answered question

2022-02-28

\sin x + \cos x = \frac{\sqrt{3} + 1}{2}

then

\tan x + \cot x =

litoshypinaylh4

Beginner2022-03-01Added 6 answers

We have

\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^{2} x + \cos^{2} x}{\sin x \cos x}

= \frac{1}{\sin x \cos x}

So you want to know how much is

\sin x \cos x

Squaring the equality you know, we have:

{(\sin x + \cos x)}^{2} = {(\frac{\sqrt{3} + 1}{2})}^{2}

\sin^{2} x + 2 \sin x \cos x + \cos^{2} x = \frac{4 + 2 \sqrt{3}}{4}

1 + 2 \sin x \cos x = 1 + \frac{\sqrt{3}}{2}

2 \sin x \cos x = \frac{\sqrt{3}}{2}

\sin x \cos x = \frac{\sqrt{3}}{4}

\tan x + \cot x = \frac{1}{\sin x \cos x} = \frac{4 \sqrt{3}}{3}

aksemaktjya

Beginner2022-03-02Added 6 answers

Another more general approach is to solve your equation for x. Since it is linear in

\sin x

and

\cos x

it can be transformed into a quadratic equation in

\tan \frac{x}{2}

\sin x + \cos x = \frac{1 + \sqrt{3}}{2} \Rightarrow \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}} + \frac{1 - \tan^{2} \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}}

= \frac{1 + \sqrt{3}}{2}

\Rightarrow 2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2} = (1 + \tan^{2} \frac{x}{2}) \frac{1 + \sqrt{3}}{2}

Set

y = \tan \frac{x}{2}

2 y + 1 - y^{2} = \frac{1 + \sqrt{3}}{2} + \frac{1 + \sqrt{3}}{2} y^{2}

\Rightarrow (1 + \frac{1 + \sqrt{3}}{2}) y^{2} - 2 y - 1 + \frac{1 + \sqrt{3}}{2} = 0

and solve for y

y_{1} = \frac{1}{3} \sqrt{3}, y_{2} = 2 - \sqrt{3}

Hence

x_{1} = 2 \arctan \frac{1}{2} \sqrt{3} = \frac{1}{3} π

x_{2} = 2 \arctan (2 - \sqrt{3}) = \frac{1}{6} π

And finally

\tan \frac{1}{3} π + \cot \frac{1}{3} π = \frac{4}{3} \sqrt{3}

\tan \frac{1}{6} π + \cot \frac{1}{6} π = \frac{4}{3} \sqrt{3}

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