Answer

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**Hint:**We can split the left hand side of the given equation. $7$ in the equation can be replaced as $4 + 3$. So we can take three as common on the LHS. Then we can apply the trigonometric relation of ${\sin ^2}x + {\cos ^2}x$. Now we can simplify the equation. Using the sine value we can find the angle and so we can find the tangent value.

**Formula used:**For angles in radians we have,

$\sin \dfrac{\pi }{6} = \dfrac{1}{2}$

$\tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$

For any angle $x$ we have, \[{\sin ^2}x + {\cos ^2}x = 1\]

**Complete step-by-step solution:**

Given that \[7{\sin ^2}x + 3{\cos ^2}x = 4\]

Since $7{\sin ^2}x = 4{\sin ^2}x + 3{\sin ^2}x$

We can write,

$\Rightarrow$\[4{\sin ^2}x + 3{\sin ^2}x + 3{\cos ^2}x = 4\]

Taking $3$ common from the second and third terms we have,

$\Rightarrow$\[4{\sin ^2}x + 3({\sin ^2}x + {\cos ^2}x) = 4\]

We know \[{\sin ^2}x + {\cos ^2}x = 1\]

Substituting this we have,

$\Rightarrow$\[4{\sin ^2}x + 3 \times 1 = 4\]

\[ \Rightarrow 4{\sin ^2}x + 3 = 4\]

Subtracting $3$ from both sides we get,

$\Rightarrow$\[4{\sin ^2}x + 3 - 3 = 4 - 3\]

\[ \Rightarrow 4{\sin ^2}x = 1\]

Dividing both sides by $4$ we get,

$\Rightarrow$\[{\sin ^2}x = \dfrac{1}{4}\]

Taking square roots on both sides we have,

$\Rightarrow$\[\sin x = \pm \dfrac{1}{2}\]

Considering positive root we have \[\sin x = \dfrac{1}{2}\]

This gives $x = \dfrac{\pi }{6}$, since $\sin \dfrac{\pi }{6} = \dfrac{1}{2}$

Using this we can find the value of $\tan x$.

So $\tan x = \tan \dfrac{\pi }{6} = \dfrac{1}{{\sqrt 3 }}$

**Hence we had proved the given statement.**

**Additional Information:**In a right angled triangle with one of the non-right angles $\theta $, $\sin \theta $ refers to the ratio of the opposite side of the angle $\theta $ to the hypotenuse and $\cos \theta $ refers to the ratio of the adjacent side of the angle $\theta $ to the hypotenuse. Whereas the $\tan \theta $ refers to the ratio of the opposite side of the angle $\theta $ to the adjacent side of the angle $\theta $.

**Note:**Actually sin function is a periodic function with period \[2\pi \]. But here we considered only one angle of sine which gives the value $\dfrac{1}{2}$. So we got the value $\dfrac{\pi }{6}$ for $x$.

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