Question

If tan x = 3/4 , π < x < 3π/2 , find the value of sin x/2 , cos x/2 and tan x/2
plz don't miss any step and any of the identity used

Answer 1

 We know that,
Sec²x = 1+tan²x
Sec²x = 1+(3/4)²
Sec²x = 1+(9/16)
Sec²x = 25/16
Secx = +- √25/16
Secx = +- 5/4

Since, π < x < 3π/2, It belongs to III quadrant in which Secx is negative

So,
Secx = -5/4
Cosx = -4/5

We have to know values of x/2
π < x < 3π/2 
then, 
π/2 < x/2 < 3π/2x2
π/2 < x/2 < 3π/4

So, x/2 belongs to II quadrant

So, Sinx/2 > 0, Cosx/2 < 0

1) 2Sin²x/2 = (1-cosx) = (1+4/5) = 9/5
      Sin²x/2 = 9/10
      Sinx/2 = +- √9/10
      Sinx/2 = +- 3/√10

    Since x/2 belongs II quad, Sinx/2 > 0
    Sinx/2 = 3/√10
    

2) 2Cos²x/2 = 1+Cosx = 1-4/5 = 1/5
      Cos²x/2 = 1/10
      Cosx/2 = +- 1/√10

   Since x/2 belongs to II quad, Cosx/2 is negative
       Cos x/2 = -1/√10

3) tanx/2 = Sinx/2 / Cos x/2
               = 3/√10 / -1/√10
               = -3
    
   Since x/2 belongs to II quad, tan x/2 is negative
   tanx/2 = -(-3)
   tanx/2 = 3

Answer 2

Answer:

Given π<x<3π2andtanx=34

π<x<3π2

⇒π2<x2<3π4→x2∈ 2nd quadrant

This means

sin(x2)→+ve

cos(x2)→−ve

tan(x2)→−ve

Now tanx=34

⇒2tan(x2)1−tan2(x2)=34

⇒8tan(x2)=3−3tan2(x2)

⇒3tan2(x2)+8tan(x2)−3=0

⇒3tan2(x2)+9tan(x2)−tan(x2)−3=0

⇒3tan(x2)(tan(x2)+3)−1(tan(x2)+3)=0

⇒(3tan(x2)−1)(tan(x2)+3)=0

This means

tan(x2)=13→not acceptable as tan(x2)→−ve

So tan(x2)→−3

Now

cos(x2)=1sec(x2)=−1√1+tan2(x2)

=−1√1+(−3)2=−1√10

Again

sin(x2)=tan(x2)×cos(x2)

=−3×(−1√10)=3√10