Question

(vii)
The value of sin (tan-'x + tan-'1/x),x>0 is​

Answer 1

Step-by-step explanation:

Given that sin

−1

(1−x)−2sin

−1

x=

2

π

let x=siny

∴sin

−1

(1−siny)−2y=

2

π

⇒sin

−1

(1−siny)=

2

π

+2y

⇒1−siny=sin(

2

π

+2y)

⇒1−siny=cos2y

⇒1−siny=1−2sin

2

y as cos2y=1−2sin

2

y

⇒2sin

2

y−siny=0

⇒2x

2

−x=0

⇒x(2x−1)=0

⇒x=0,2x−1=0

⇒x=0,2x=1

⇒x=0,x=

2

1

But x=

2

1

does not satisfy the given equation

∴x=0 is the solution of the given equation.