Question

If $\tan x=\dfrac{3}{4}$ , x lies in the third quadrant, find the values of $\sin \dfrac{x}{2},\text{ cos}\dfrac{x}{2}\text{ and }\tan \dfrac{x}{2}$ .

Answer 1

Hint: Start by using the formula that $\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}$ , which would give you a quadratic equation in terms of $\tan \dfrac{x}{2}$ . Now once you have got the value of $\tan \dfrac{x}{2}$ from the quadratic equation, you can easily find other trigonometric ratios using the relation between the trigonometric ratios.

Complete step-by-step answer:

We will start with the solution to the above question by finding the value of $\tan \dfrac{x}{2}$ .
We know that $\tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}$ . So, if we use the formula for tanx, we get
$\tan x=\dfrac{2\tan \dfrac{x}{2}}{1-{{\tan }^{2}}\dfrac{x}{2}}$
Now we will put the value of tanx from the question. On doing so, we get
$\dfrac{3}{4}=\dfrac{2\tan \dfrac{x}{2}}{1-{{\tan }^{2}}\dfrac{x}{2}}$
$\Rightarrow 3-3{{\tan }^{2}}\dfrac{x}{2}=8\tan \dfrac{x}{2}$
$\Rightarrow 3{{\tan }^{2}}\dfrac{x}{2}+8\tan \dfrac{x}{2}-3=0$
$\Rightarrow 3{{\tan }^{2}}\dfrac{x}{2}+9\tan \dfrac{x}{2}-\tan \dfrac{x}{2}-3=0$
$\Rightarrow 3\tan \dfrac{x}{2}\left( \tan \dfrac{x}{2}+3 \right)-1\left( \tan \dfrac{x}{2}+3 \right)=0$
$\Rightarrow \left( 3\tan \dfrac{x}{2}-1 \right)\left( \tan \dfrac{x}{2}+3 \right)=0$
Therefore, $\tan \dfrac{x}{2}=\dfrac{1}{3}\text{ or -3}$ .
It is given that x lies in the third quadrant. Then we can say that $\dfrac{x}{2}$ will for sure lie in the second quadrant and tangent is negative in the second quadrant, so the value of $\tan \dfrac{x}{2}=-3$ .
We know that ${{\sec }^{2}}\dfrac{x}{2}=1+{{\tan }^{2}}\dfrac{x}{2}.$ So, if we put the value of $\tan \dfrac{x}{2}$ in the formula, we get
$se{{c}^{2}}x=1+{{\left( -3 \right)}^{2}}$
$\Rightarrow se{{c}^{2}}x=1+9$
$\Rightarrow {{\sec }^{2}}x=10$
Now we know that ${{a}^{2}}=b$ implies $a=\pm \sqrt{b}$ . So, our equation becomes:
$\Rightarrow \sec x=\pm \sqrt{10}$
Now, as we know that $\dfrac{x}{2}$ lies in the second quadrant and $\sec \dfrac{x}{2}$ is negative in the second quadrant.
$\therefore \sec \dfrac{x}{2}=-\sqrt{10}$
Now we know that cosx is inverse of secx.
$\therefore \cos \dfrac{x}{2}=\dfrac{1}{\sec \dfrac{x}{2}}=-\dfrac{1}{\sqrt{10}}$
Now using the property that tanx is the ratio of sinx to cosx, we get
$\tan \dfrac{x}{2}=\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}$
$\Rightarrow \tan \dfrac{x}{2}\cos \dfrac{x}{2}=\sin \dfrac{x}{2}$
$\Rightarrow \sin \dfrac{x}{2}=-3\times \left( -\dfrac{1}{\sqrt{10}} \right)=\dfrac{3}{\sqrt{10}}$

Note: It is useful to remember the graph of the trigonometric ratios along with the signs of their values in different quadrants. For example: sine is always positive in the first and the second quadrant while negative in the other two. Also, you need to remember the properties related to complementary angles and trigonometric ratios. As you saw in the above solution, we had used the result that $\dfrac{x}{2}$ will for sure lie in the second quadrant. We arrived at this result as follows:
As we knew that $x$ lies in the second quadrant, we can say:
$\pi \le x\le \dfrac{3\pi }{2}$
Now if we divide each term in the inequality by 2, we get
$\dfrac{\pi }{2}\le \dfrac{x}{2}\le \dfrac{3\pi }{4}$
Using this result we can say that $\dfrac{x}{2}$ lies in the second quadrant.
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