Question

1-13
S
sin-cos/sin+cos=1-√3/1+√3​

Answer 1

Answer:

Solution:

sin−1[cos(sin−1x)]+cos−1[sin(cos−1x)]

=sin−1[sin(π2−sin−1x)]+cos−1[cos(π2−cos−1x)]

=π2−sin−1x+π2−cos−1x

=π−(sin−1x+cos−1x)

=π−π2

=π2

Questions from Inverse Trigonometric Functions

Step-by-step explanation:

hope it helps you.