tanx=−43, x in quadrant II. Find the value of sinx2,cosx2,tanx2
As tanx=−43,π2<x<π
i.e x lies in 2nd quadrant
Hence tanx=−43⇒sinx=4√42+32=45
And cosx=−5√42+32=−35
Now using 1−cosx=2sin2x2⇒sinx2=±√1−cosx2, we get
sinx2=±
⎷1−(−35)2=±√810
As π2<x<π⇒π4<x2<π2 and sine is positive in 1st quadrant
Then sinx2=2√5
Using 1+cosx=2cos2x2⇒cosx2=±√1+cosx2
We get, cosx2=±
⎷1+(−35)2=±√210
π2<x<π⇒π4<x2<π2 and cos is positive 1st quadrant
∴cosx2=1√5
Using cosx=1−tan2x21+tan2x2⇒tanx2=±√1−cosx1+cosx
We get, tanx2=±
⎷1−(−35)1+(−35)=±√4
As π2<x<π⇒π4<x2<π2 and tan is positive in 1st quadrant
∴tanx2=2