Question

If \[7{\sin ^2}x + 3{\cos ^2}x = 4\], show that $\tan x = \dfrac{1}{{\sqrt 3 }}$.

Answer 1

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 $x$.