Find `sin``x/2``,cos``x/2`and `tan``x/2`of the following : `tanx=-4/3,x`in quadrant II

Explanation

`therefore tanx=-4/3`
`therefore (2tanx/2)/(1-tan^(2)x/2)=-4/3`
or `-4+4tan^(2)x/2=6tanx/2`
or `2tan^(2)x/2-3tanx/2-2=0`
or `2tanx/2(tanx/2-2)+1(tanx/2-2)=0`
then `(2tanx/2+1=0)` or `(tanx/2-2)=0`
If `2tanx/2+1=0`, then `tanx/2=-1/2`,
and if `tanx/2-1=0`, then `tanx/2=2`
But x lies in second quadrant from which sin x positive, so the value `-1/2` of `tanx/2` is not acceptable.
Then `tanx/2=2`
So, `cosx=(1-tan^(2)x/2)/(1+tan^(2)x/2) = (1-(2)^(2))/(1+(2)^(2))=-3/5`
`therefore cosx=-3/5`
`rArr 1-2sin^(2)x/2 = -3/5` or `2cos^(2)x/2-1=-3/5`
If `1-2sin^(2)x/2=-3/5`
`2sin^(2)x/2=1+3/5=8/5` or `sin^(2)x/2=4/5`
or `sinx/2=2/sqrt(5)`
If `2cos^(2)x/2-1=-3/5`
then `2cos^(2)x/2-1=-3/5`
then `2cos^(2)x/2=1-3/5=2/5` or `cos^(2)x/2=1/5`
or `cosx/2=1/sqrt(5)`
Therefore, `sinx/2=2/sqrt(5), cosx/2=1/5`
and `tanx/2=2` Ans.