Statement-1: {cot ^{ - 1}}left[ {frac{{log (e | vedclass

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(D) Let $y = \log(x^2)$. Then $\log(e/x^2) = \log e - \log x^2 = 1 - y$ and $\log(ex^2) = \log e + \log x^2 = 1 + y$.Statement-$2$: The expression is

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is the correct explanation of Statement-$1$.

Vedclass · Answer

(D) Let $y = \log(x^2)$. Then $\log(e/x^2) = \log e - \log x^2 = 1 - y$ and $\log(ex^2) = \log e + \log x^2 = 1 + y$.Statement-$2$: The expression is ${	an ^{ - 1}}\left[ {\frac{{1 + y}}{{1 - y}}} ight]$. Using the formula ${	an ^{ - 1}}A + {	an ^{ - 1}}B = {	an ^{ - 1}}\left( {\frac{{A + B}}{{1 - AB}}} ight)$,we have ${	an ^{ - 1}}1 + {	an ^{ - 1}}y = {	an ^{ - 1}}\left( {\frac{{1 + y}}{{1 - 1 \cdot y}}} ight) = {	an ^{ - 1}}\left( {\frac{{1 + y}}{{1 - y}}} ight)$. Thus,Statement-$2$ is true.Statement-$1$: Let $u = \frac{1-y}{1+y}$. The expression is ${\cot ^{ - 1}}(u) + {\cot ^{ - 1}}(1/u)$. We know that ${\cot ^{ - 1}}(u) = {	an ^{ - 1}}(1/u)$ for $u > 0$. If $u > 0$,then ${	an ^{ - 1}}(1/u) + {	an ^{ - 1}}(u) = \frac{\pi}{2}$. If $u < 0$,${\cot ^{ - 1}}(u) = \pi + {	an ^{ - 1}}(1/u)$,leading to a different result. However,for the identity to hold generally as $\frac{\pi}{2}$,the arguments must be positive. Since Statement-$2$ provides the decomposition,it acts as the basis for the simplification of the terms in Statement-$1$. Thus,Statement-$1$ is true and Statement-$2$ is the correct explanation.

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