Question

if sec x =2 and π<x<3π/2 then sin x = ?​

Answer 1

Answer:

Step-by-step explanation:

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