yasusar0

2022-07-18

Solving inverse trig question without forming cases

Question:

Find $x$ if $\mathrm{arctan}(x+3)-\mathrm{arctan}(x-3)=\mathrm{arctan}(3/4)$

My attempt:

I know the formula:

$\mathrm{arctan}x-\mathrm{arctan}y=\mathrm{arctan}(\frac{x-y}{1+xy})$

for both $x,y>0$

If both $x,y$ are NOT greater than $0$ then we'll need to form a second case in the above formula.

I solved assuming both $x+3$ and $x-3$ are greater than zero and got a quadratic in the end solving which I got $x=\pm 4$

I dutifully rejected $x=-4$ as it invalidated my assumption. However on putting in the above equation I found that it DOES satisfy the equation.

I do not wish to form cases for positive/negative. I wish to know if there is a simpler and direct way to solve such a question. Thanks!

Question:

Find $x$ if $\mathrm{arctan}(x+3)-\mathrm{arctan}(x-3)=\mathrm{arctan}(3/4)$

My attempt:

I know the formula:

$\mathrm{arctan}x-\mathrm{arctan}y=\mathrm{arctan}(\frac{x-y}{1+xy})$

for both $x,y>0$

If both $x,y$ are NOT greater than $0$ then we'll need to form a second case in the above formula.

I solved assuming both $x+3$ and $x-3$ are greater than zero and got a quadratic in the end solving which I got $x=\pm 4$

I dutifully rejected $x=-4$ as it invalidated my assumption. However on putting in the above equation I found that it DOES satisfy the equation.

I do not wish to form cases for positive/negative. I wish to know if there is a simpler and direct way to solve such a question. Thanks!

Vartavk

Beginner2022-07-19Added 11 answers

Your answer $\{4,-4\}$ is right.

A full solution:

Since $\mathrm{tan}(\alpha -\beta )=\frac{\mathrm{tan}\alpha -\mathrm{tan}\beta}{1+\mathrm{tan}\alpha \mathrm{tan}\beta}$ and $\mathrm{tan}\mathrm{arctan}x=x$ for all real $x$, we obtain

$\mathrm{tan}(\mathrm{arctan}(x+3)-\mathrm{arctan}(x-3))=\mathrm{tan}\mathrm{arctan}\frac{3}{4}$

or

$\frac{x+3-(x-3)}{1+(x+3)(x-3)}=\frac{3}{4}$

or

${x}^{2}=16,$

which gives $x=4$ or $x=-4$

The checking of these numbers gets that indeed, they are roots and we are done!

A full solution:

Since $\mathrm{tan}(\alpha -\beta )=\frac{\mathrm{tan}\alpha -\mathrm{tan}\beta}{1+\mathrm{tan}\alpha \mathrm{tan}\beta}$ and $\mathrm{tan}\mathrm{arctan}x=x$ for all real $x$, we obtain

$\mathrm{tan}(\mathrm{arctan}(x+3)-\mathrm{arctan}(x-3))=\mathrm{tan}\mathrm{arctan}\frac{3}{4}$

or

$\frac{x+3-(x-3)}{1+(x+3)(x-3)}=\frac{3}{4}$

or

${x}^{2}=16,$

which gives $x=4$ or $x=-4$

The checking of these numbers gets that indeed, they are roots and we are done!

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