MHB How Do You Solve tan(2x - 5) = cot(x + 5) in the Interval 0 < x < 90?

[Monoxdifly]

If 0 < x < 90, what is the solution of tan(2x - 5) = cot(x + 5)?
I got stuck in tan(2x - 5)tan(x + 5) = 1. What should I do after that?

 

[I like Serena]

Suppose we substitute $\cot\alpha=\tan(\frac\pi 2-\alpha)$ and apply $\arctan$? (Wondering)

 

[skeeter]

another approach ...

$\tan{\alpha} = \cot{\beta}$

$\dfrac{\sin{\alpha}}{\cos{\alpha}} = \dfrac{\cos{\beta}}{\sin{\beta}}$

$\cos{\alpha} \cdot \cos{\beta} = \sin{\alpha} \cdot \sin{\beta}$

$\cos{\alpha} \cdot \cos{\beta} - \sin{\alpha} \cdot \sin{\beta} = 0$

$\cos(\alpha + \beta) = 0$

 

[Monoxdifly]

another approach ...

$\tan{\alpha} = \cot{\beta}$

$\dfrac{\sin{\alpha}}{\cos{\alpha}} = \dfrac{\cos{\beta}}{\sin{\beta}}$

$\cos{\alpha} \cdot \cos{\beta} = \sin{\alpha} \cdot \sin{\beta}$

$\cos{\alpha} \cdot \cos{\beta} - \sin{\alpha} \cdot \sin{\beta} = 0$

$\cos(\alpha + \beta) = 0$

I like this other approach since I can comprehend it easier. I got $$30^{\circ}$$ this way. Is that right?

 

[skeeter]

I like this other approach since I can comprehend it easier. I got $$30^{\circ}$$ this way. Is that right?

substitute your solution into the original equation ...

is $\tan(55^\circ) = \cot(35^\circ)$ ?

 

[Monoxdifly]

substitute your solution into the original equation ...

is $\tan(55^\circ) = \cot(35^\circ)$ ?

I don't know, I can't count that without using a calculator or manually measuring the ratios with ruler and protactor.
Oh wait, I can just draw a random right triangle and label the angles.
Oh yes, $\tan(55^\circ) = \cot(35^\circ)$.

 

[skeeter]

I don't know ...

trig functions and cofunctions