Q) $$ an (\phi+ heta)=3 $$ and $$ an \phi+ an heta=1 $$ Find the value of $$ \phi $$ and $$ heta $$.

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Q) $$ 	an (\phi+	heta)=3 $$ and $$ 	an \phi+	an 	heta=1 $$ Find the value of $$ \phi $$ and $$ 	heta $$.

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To solve the problem, we start with the given equations:

1. $$ \tan(\phi + \theta) = 3 $$
2. $$ \tan \phi + \tan \theta = 1 $$

We can use the tangent addition formula for the first equation:

$$
\tan(\phi + \theta) = \frac{\tan \phi + \tan \theta}{1 - \tan \phi \tan \theta}
$$

Substituting the second equation into the tangent addition formula gives us:

$$
3 = \frac{1}{1 - \tan \phi \tan \theta}
$$

Cross-multiplying yields:

$$
3(1 - \tan \phi \tan \theta) = 1
$$

Expanding this, we have:

$$
3 - 3\tan \phi \tan \theta = 1
$$

Rearranging gives:

$$
3\tan \phi \tan \theta = 2
$$

Thus, we can express the product of the tangents:

$$
\tan \phi \tan \theta = \frac{2}{3}
$$

Now we have two equations:

1. $$ \tan \phi + \tan \theta = 1 $$
2. $$ \tan \phi \tan \theta = \frac{2}{3} $$

Let $$ x = \tan \phi $$ and $$ y = \tan \theta $$. We can rewrite the equations as:

1. $$ x + y = 1 $$
2. $$ xy = \frac{2}{3} $$

From the first equation, we can express $$ y $$ in terms of $$ x $$:

$$
y = 1 - x
$$

Substituting this into the second equation gives:

$$
x(1 - x) = \frac{2}{3}
$$

Expanding this, we have:

$$
x - x^2 = \frac{2}{3}
$$

Rearranging leads to a quadratic equation:

$$
x^2 - x + \frac{2}{3} = 0
$$

To eliminate the fraction, we can multiply the entire equation by 3:

$$
3x^2 - 3x + 2 = 0
$$

Now we can use the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$:

Here, $$ a = 3 $$, $$ b = -3 $$, and $$ c = 2 $$:

$$
x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3}
$$
$$
x = \frac{3 \pm \sqrt{9 - 24}}{6}
$$
$$
x = \frac{3 \pm \sqrt{-15}}{6}
$$

Since the discriminant is negative, we have complex solutions. We can express the square root of $$-15$$ as $$ i\sqrt{15} $$:

$$
x = \frac{3 \pm i\sqrt{15}}{6}
$$
$$
x = \frac{1}{2} \pm \frac{i\sqrt{15}}{6}
$$

Thus, we have:

$$
\tan \phi = \frac{1}{2} + \frac{i\sqrt{15}}{6}, \quad \tan \theta = \frac{1}{2} - \frac{i\sqrt{15}}{6}
$$

Now we can find $$ \phi $$ and $$ \theta $$ using the arctangent function:

$$
\phi = \tan^{-1}\left(\frac{1}{2} + \frac{i\sqrt{15}}{6}\right), \quad \theta = \tan^{-1}\left(\frac{1}{2} - \frac{i\sqrt{15}}{6}\right)
$$

Thus, the values of $$ \phi $$ and $$ \theta $$ are complex numbers. The final answer is:

$$
\phi = \tan^{-1}\left(\frac{1}{2} + \frac{i\sqrt{15}}{6}\right), \quad \theta = \tan^{-1}\left(\frac{1}{2} - \frac{i\sqrt{15}}{6}\right)
$$
The values of $$ \phi $$ and $$ \theta $$ are complex numbers. $$ \phi = \tan^{-1}\left(\frac{1}{2} + \frac{i\sqrt{15}}{6}\right), \quad \theta = \tan^{-1}\left(\frac{1}{2} - \frac{i\sqrt{15}}{6}\right) $$

Q) \( \tan (\phi+\theta)=3 \) and \( \tan \phi+\tan \theta=1 \) Find the value of \( \phi \) and \( \theta \).

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