CUET Preparation Today
If -1 < x < 0 , then sin-1x equals |
$\pi-\cos^{-1}(\sqrt{1-x^2})$ $\tan^{-1}\frac{x}{\sqrt{1-x^2}}$ $-\cot^{-1}(\frac{\sqrt{1-x^2}}{x})$ cosec-1x |
$\tan^{-1}\frac{x}{\sqrt{1-x^2}}$ |
Since -1< x < 0, -π/2 < sin–1x < 0. Let sin-1 x = α i.e. sinα = x Then $tanα=\frac{x}{\sqrt{1-x^2}}⇒ α = tan^{-1}\frac{x}{\sqrt{1-x^2}}⇒ sin^{-1}x = tan^{-1}\frac{x}{\sqrt{1-x^2}}$. Hence (B) is the correct answer. Alternative: $sin^{-1}x=-sin^{-1}(-x)=-tan^{-1}(-\frac{x}{\sqrt{1-x^2}})=tan^{-1}\frac{x}{\sqrt{1-x^2}}$ |
