Question

Derivative of `sin^-1 ("t"/sqrt(1 + "t"^2))` with respect to `cos^-1 (1/sqrt(1 + "t"^2))` is ______.

पर्याय

• 1

• cot t

• tan t

• 0

Answer 1

Derivative of `sin^-1 ("t"/sqrt(1 + "t"^2))` with respect to `cos^-1 (1/sqrt(1 + "t"^2))` is 1.

Explanation:

Let y = `sin^-1 ("t"/sqrt(1 + "t"^2))`

Put t = tan θ ⇒ θ = tan^-1 t

`= sin^-1((tan theta)/sqrt(1 + tan^2 theta)) = sin^-1 ((tan theta)/(sec theta))`

= sin^-1(sin θ) = θ = tan^-1 t

and z = `cos^-1 (1/sqrt(1 + "t"^2)) = cos^-1(1/(sqrt(1 + tan^2theta)))`

= cos^-1 (cos θ)

= θ = tan^-1t

`therefore "dy"/"dx" = ("dy"/"dt")/("dz"/"dt")`

`= (1/((1 + "t"^2)))/(1/((1 + "t"^2)))` = 1