Question

find sin^-1(√[1-x^2])​

Answer 1

Answer:

REF.Image.

I=sin

−1

x+sin

−1

1−x

2

In △ABC

⇒sinA=(

1

x

)

⇒A=sin

−1

(x)

sinC=

1

1−x

2

⇒C=sin

−1

(

1−x

2

)

∴A+C=sin

−1

(x)+sin

−1

(

1−x

2

)

A+B+C=180

∘

⇒

A+90+C=180

∘

⇒A+C=90

∘

=π/2

∴sin

−1

(x)+sin

−1

(

1−x

2

)=π/2

Thank you and have a nice day