Question

Given, tanx = - 4/3. Lies in 2nd quadrant. Find sinx/2, cosx/2 and tanx/2.

Answer 1

Use the identity,

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

Here,

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

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

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

Answer 2

Answer:

Use the identity,

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

Here,

 \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}    

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

Hence the solution is  \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}}  

Step-by-step explanation: