Question

Assertion :Derivative of $si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$ with respect to $co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ is $1$ for $0<x<1.$ Reason: $si{n}^{-1}\left(\frac{2x}{1+x^2}\right)=co{s}^{-1}\left(\frac{1-x^2}{1+x^2}\right)$ for $-1\le x\le 1$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$\frac{d}{dx}\left({\sin }^{-1}\left(\frac{2x}{1+x^2}\right)\right)=\frac{\frac{d}{dx}\left(\frac{2x}{1+x^2}\right)}{\sqrt{1-\frac{{4x}^2}{{\left({1+x}^2\right)}^2}}}$
$=\frac{(x^2+1)\frac{d}{dx}\left(x\right)-x\frac{d}{dx}\left({1+x}^2\right)}{{(x^2+1)}^2}\times \frac{2}{\sqrt{1-\frac{{4x}^2}{{(1+x^2)}^2}}}$
$=\frac{2(1+x^2-2x^2)}{{(1+x^2)}^2\sqrt{1-\frac{{4x}^2}{{(1+x^2)}^2}}}=\frac{2}{(x^2+1)}$ for $(0<x<1)$
$\frac{d}{dx}\left({\cos }^{-1}\left(\frac{{1-x}^2}{1+x^2}\right)\right)=-\frac{\frac{d}{dx}\left(\frac{1-x^2}{1+x^2}\right)}{\sqrt{1-{\left(\frac{{1-x}^2}{1+x^2}\right)}^2}}$
$=-\frac{-1}{\sqrt{1-{\left(\frac{1-x^2}{1+x^2}\right)}^2}}\times \frac{-\left({1-x}^2\right)\frac{d}{dx}(1+x^2)+(1+x^2)\frac{d}{dx}(1-x^2)}{{(1+x^2)}^2}$
$=\frac{2x(1+x^2)-\left({1-x}^2\right)2x}{(1+x^2)\sqrt{1-{\left(\frac{1-x^2}{1+x^2}\right)}^2}}=\frac{2\sqrt{\frac{x^2}{{(1+x^2)}^2}}}{x}=\frac{2}{(x^2+1)}$
Reason
${\sin }^{-1}\left(\frac{2x}{{1+x}^2}\right)={\cos }^{-1}\left(\sqrt{1-{\left(\frac{2x}{1+x^2}\right)}^2}\right)$
$={\cos }^{-1}\left(\sqrt{\frac{1+x^4+{2x}^2-{4x}^2}{{(1+x^2)}^2}}\right)={\cos }^{-1}\left(\sqrt{\frac{{(1-x^2)}^2}{{(1+x^2)}^2}}\right)$
$={\cos }^{-1}\frac{(1-x^2)}{(1+x^2)}$