We have, tanx = −

4

3

, x in quadrant II
since , x in quadrant II ⇒

π

2

< x < π ,
⇒

π

4

<

x

2

<

π

2

⇒

x

2

lies in 1st quadrant ⇒ sin

x

2

> 0 , cos

x

2

> 0 , tan

x

2

> 0
Also 1 + tan2x = sec2x ⇒ sec2x ⇒ sec2x = 1 +

16

9

=

25

9

⇒ sec x = ±

5

3

⇒ cos x = - 3/5
Since

π

2

< x < π , ∴ cos x is - ve
Now cos

x

2

= ± √

1+cosx

2

= √

1− 3 5

2

Since cos

x

2

is + ve
= √

2

5

×

1

2

= √

1

5

=

1

√5

sin

x

2

= ± √

1−cosx

2

= ± √

1+ 3 5

2

= √

8

5

×

1

2

=

2

√5

Since sin

x

2

> 0
tan

x

2

= sin

x

2

cos

x

2

=

2

√5

×

√5

1

= 2
Hence sin

x

2

=

2√5

2

, cos

x

2

=

√5

2

, tan

x

2

= 2