Question

If `sin^-1"x" + tan^-1"x" = pi/2`, prove that `2"x"^2 + 1 = sqrt5`  

Answer 1

`sin ^-1"x"+tan^-1"x"=pi/2`

⇒   `tan^-1("x"/ sqrt (1-"x"^2))  + tan^-1"x" =pi/2`

 

⇒   `tan^-1[("x"/ sqrt (1-"x"^2)+ "x")/(1 -"x"/(sqrt(1-"x"^2)). "x")] = pi/2`

 

⇒   `tan^-1(("x"+"x"sqrt(1-"x"^2))/(sqrt(1-"x"^2)-"x"^2)) = pi/2`

 

⇒   `("x"+"x"sqrt(1-"x"^2))/(sqrt(1-"x"^2)-"x"^2 )= tan(pi/2)`

 

⇒   `(sqrt(1-"x"^2)-"x"^2)/("x"+"x"sqrt(1-"x"^2)) = cot (pi/2)=0`

 

⇒ `sqrt(1-"x"^2) - "x"^2 = 0`

⇒   `"x"^2= sqrt(1-"x"^2)`

⇒   `"x"^4 = 1-"x"^2`

⇒  `"x"^4 +"x"^2 - 1 = 0`

⇒ let `"x"^2 ="t"`

⇒ t² + t - 1 = 0

t = `(-1±sqrt(1+4))/2`

t = `(-1±sqrt(5))/2 =⇒ 2"t" = -1± sqrt5`

⇒  ` 2"x"^2+1 = ± sqrt5`   ignoring negative value

We get : `2"x"^2+1 = sqrt5`  = proved