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Discussion Forum



1)

sin−1√32+sin−1√23= 


A) sin−1√3+√22√3

B) π−sin−1(√3+√22√3)

C) −π−sin−1(√3+√22√3)

D) π+sin−1(√3+√22√3)

Answer:

Option B

Explanation:

We have, 

sin−1√32+sin−1√23

         =π−sin−1[√32√1−(√23)2+√32√1−(√32)2]

 As    0< √32=0.866,√23=0.816≤ 1   and

  (√32)2+(√23)2=1.4167>1

                   [∵sin−1x+sin−1y=π−sin−1

                 x−√1−y2+y√1−x2,if0<x,y≤1    and x2+y2>1]

    = π−sin−1[√32√1−23+√23√1−34]

   = π−sin−1[√32√3−23+√23√4−34]

      =π−sin−1[√321√3+√2√312]

  =π−sin−1[√3+√22√3]