Question

\[\sin^{- 1} \sqrt{1 - x^2}\] with respect to \[\cot^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right),\text { if }0 < x < 1\] ? 

Answer 1

\[\text { Let, u }= \sin^{- 1} \left( \sqrt{1 - x^2} \right)\]

\[\text {Put x } = \cos\theta\]

\[ \Rightarrow \theta = \cos^{- 1} x\]

\[\text{We get, }u = \sin^{- 1} \left( \sin\theta \right) ...... \left( i \right)\]

\[\text {  Let, v } = co t^{- 1} \left( \frac{x}{\sqrt{1 - x^2}} \right)\]

\[ \Rightarrow v = co t^{- 1} \left( \frac{\cos\theta}{\sqrt{1 - \cos^2 \theta}} \right) \]

\[ \Rightarrow v = co t^{- 1} \left( \frac{\cos\theta}{\sin\theta} \right)\]

\[ \Rightarrow v = co t^{- 1} \left( cot\theta \right) . . . \left( ii \right)\]

\[\text { Here }, \]

\[ 0 < x < 1\]

\[ \Rightarrow 0 < \cos\theta < 1\]

\[ \Rightarrow 0 < \theta < \frac{\pi}{2}\]

\[\text { So, from equation} \left( i \right), \]

\[u = \theta \left[ \text { Since,} \sin^{- 1} \left( \sin\theta \right) = \theta, \text { if }\theta \in \left[ - \frac{\pi}{2}, \frac{\pi}{2} \right] \right]\]

\[ \Rightarrow u = \cos^{- 1} x\]

Differentiating it with respect to x,

\[\frac{du}{dx} = \frac{- 1}{\sqrt{1 - x^2}} . . . \left( iii \right)\]

\[\text { From equation } \left( ii \right), \]

\[v = \theta \left[ \text {Since}, co t^{- 1} \left( cot\theta \right) = \theta, \text { if }\theta \in \left( 0, \pi \right) \right]\]

\[ \Rightarrow v = \cos^{- 1} x\]

Differentiating it with respect to x,

\[\frac{dv}{dx} = \frac{- 1}{\sqrt{1 - x^2}} . . . \left( iv \right)\]

\[\text { Dividing equation } \left( iii \right) \text { by } \left( iv \right), \]

\[\frac{\frac{du}{dx}}{\frac{dv}{dx}} = \left( \frac{- 1}{\sqrt{1 - x^2}} \right)\left( \frac{\sqrt{1 - x^2}}{- 1} \right)\]

\[ \therefore \frac{du}{dv} = 1\]