Question

Trignometry question

Answer 1

\(\large \color{black }{\begin{align}
\tan ^{-1}y=\tan ^{-1}x+\tan ^{-1}\dfrac{2x}{1-x^2}\hspace{.33em}\\~\\
\end{align}}\)

where

\(\large \color{black }{\begin{align}
\mid x \mid <\dfrac{1}{\sqrt 3}\hspace{.33em}\\~\\
\end{align}}\)

Find the value of \(\Large y\)

Answer 2

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Answer 3

\[\huge \bf \tan^{-1}\frac{2x}{1-x^2}=\tan^{-1}\frac{x+x}{1-(x)(x)}\]

Answer 4

so,
\[\huge \bf \tan^{-1}\frac{2x}{1-x^2}=\tan^{-1}x+\tan^{-1}x\]

Answer 5

just plug in the question, and solve for y ? @mathmath333

Answer 6

Refer to the Mathematica calculation attached.

Answer 7

its so long and freaky

Answer 8

hint:
if I compute the tangent of both sides, I get this:
\[\Large y = \frac{{x + \frac{x}{{1 - {x^2}}}}}{{1 - \frac{{{x^2}}}{{1 - {x^2}}}}} = ...?\]

Answer 9

how did u get the above form @Michele_Laino

Answer 10

I have compute the tangent function of both sides of your equation, as I wrote above

Answer 11

computed*

Answer 12

can u send me the link for formula for inverse trig

Answer 13

I have used this formula:
\[\begin{gathered}
\alpha = \beta + \gamma \hfill \\
\tan \alpha = \tan \left( {\beta + \gamma } \right) = \frac{{\tan \beta + \tan \gamma }}{{1 - \tan \beta \cdot \tan \gamma }} \hfill \\
\end{gathered} \]

Answer 14

where:
\[\begin{gathered}
\alpha = \arctan y \hfill \\
\beta = \arctan x \hfill \\
\gamma = \arctan \left( {\frac{{2x}}{{1 - {x^2}}}} \right) \hfill \\
\end{gathered} \]

Answer 15

What does the problem mean by: "Find the values of \(y\) "? It looks like you have to take the tangent of both sides of the equal sign, then apply the angle sum identity and use algebra to simplify. Note that you must make note of the domain originally specified as well.

Note: @Michele_Laino had the right idea, but there is a small typo in his expression for \(y\). There should be a \(\color{red}2x^2\) in the denominator.

\[
y = \Large \frac{{x + \frac{x}{1 - x^2}}}{{1 - \frac{{{\color{red}2x^2}}}{1 - x^2}}}
\]

This simplifies to \[y=\frac{x(3-x^2)}{1-3x^3}\]

The graph is here: https://www.desmos.com/calculator/lauavrvkdr

Answer 16

Another note: @robtobey 's mathematica sheet graphs the (correct) function but only from \(\color{red}0\) to \(\frac{1}{\sqrt{3}}\), but it should be graphed over this interval: \[\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right).\]

Answer 17

thanks for your comment! @mathteacher1729