Question

If Tan x = (-4 / 3) , where x is in second quadrant , then find Sin (x/2), Cos (x/2) and Tan (x/2).

Answer 1

Answer:

Use the identity,

\tan x=\frac{2\tan (x/2)}{1-\tan ^2(x/2)}tanx=

1−tan

2

(x/2)

2tan(x/2)

.

Here,

$$\begin{lgathered}\tan x=\frac{2\tan (x/2)}{1-\tan ^2(x/2)} =-\frac{4}{3} \\ 4\tan ^2(x/2)-6\tan (x/2)-4=0\\ 2\tan ^2(x/2)-3\tan (x/2)-2=0\\ (\tan (x/2)-2)(2\tan (x/2)+1)=0\\ \tan (x/2)=2,-\frac{1}{2}\end{lgathered}$$

Since $$x$$ lies in the 2nd quadrant, $$x/2$$ lies in the first quadrant. So, $$\tan (x/2)>0$$ .

Hence the solution is $$\begin{lgathered}\tan (x/2)=2\\ \sin (x/2)=\frac{2}{\sqrt{2^2+1}} =\frac{2}{\sqrt{5}} \\ \sin (x/2)=\frac{1}{\sqrt{2^2+1}} =\frac{1}{\sqrt{5}}\end{lgathered}$$

Step-by-step explanation:

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