how do i take the argument out of the arctg and solve this simple equation

[tommycashmoney]

if tan = sin(x)/cos(x), arctg(x) = tan^-1(x), that said
this is the equation i need to solve
-arctg(x/12) -arctg(x/10) = 93

i believe that
arctg(x) = y, is the same as x = tan(y), (not very sure about this formula, but if it is true i don't know how can i apply it with two arctg in the same equation.

so what would be the right process or the right formula to solve it

 

[Dr.Peterson]

if tan = sin(x)/cos(x), arctg(x) = tan^-1(x), that said
this is the equation i need to solve
-arctg(x/12) -arctg(x/10) = 93

i believe that
arctg(x) = y, is the same as x = tan(y), (not very sure about this formula, but if it is true i don't know how can i apply it with two arctg in the same equation.

so what would be the right process or the right formula to solve it

What you've said is right, so far, though you also have to keep the range of arctg in mind.

But the equation seems wrong to me; it says that the sum of two arctangents, each of which must be between -pi/2 and pi/2 is -93. That's impossible. Are you expressing angles in degrees (which is not normal in such equations), or is something typed incorrectly?

I haven't tried solving this yet, but my first thought is to define, say, u = arctg(x/12) and v = arctg(x/10), and write the equation in terms of u and v, then take the tangent of both sides and hope something useful happens.

 

[tommycashmoney]

What you've said is right, so far, though you also have to keep the range of arctg in mind.

But the equation seems wrong to me; it says that the sum of two arctangents, each of which must be between -pi/2 and pi/2 is -93. That's impossible. Are you expressing angles in degrees (which is not normal in such equations), or is something typed incorrectly?

I haven't tried solving this yet, but my first thought is to define, say, u = arctg(x/12) and v = arctg(x/10), and write the equation in terms of u and v, then take the tangent of both sides and hope something useful happens.

ye, i was expressing angles in degrees, i should have clarified it, i will try, but it's not promising

 

[Deleted member 4993]

if tan = sin(x)/cos(x), arctg(x) = tan^-1(x), that said
this is the equation i need to solve
-arctg(x/12) -arctg(x/10) = 93

i believe that
arctg(x) = y, is the same as x = tan(y), (not very sure about this formula, but if it is true i don't know how can i apply it with two arctg in the same equation.

so what would be the right process or the right formula to solve it

You say:

if tan = sin(x)/cos(x), ..................................... that is incorrect. It should be ........ if tan(x) = sin(x)/cos(x),

Do not start your solution with a belief ( → i believe that). Instead show your knowledge (→ I know that).

You should start this problem with the following:

arctg(x/10) = y →​

​

x/10 = ? →​

​

x/12 = ? →​

​

arctg(x/12) = ?​

and continue....

 

[Dr.Peterson]

ye, i was expressing angles in degrees, i should have clarified it, i will try, but it's not promising

The method I suggested works nicely. You will also find it helpful at some point to make a substitution like k = -tan(93 deg) so you don't have to write that number out repeatedly.

By the way, there is rarely one "right process"; I avoid even saying "you should". But as I implied, trying something without being sure it will work is normal procedure! You can't be sure what is "promising" until you try it. Be optimistic. Be bold. Be willing to try something and then try something else.