If cos ^{-1}left(frac{y}{b}right)=n log left( | vedclass

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(B) Given,$\cos ^{-1}\left(\frac{y}{b}\right)=n \log \left(\frac{x}{n}\right)$.Differentiating both sides with respect to $x$:$-\frac{1}{\sqrt{1-(y/b)

Vedclass · Accepted Answer

$x y_{1}+n \sqrt{b^{2}-y^{2}}=0$

Vedclass · Answer

(B) Given,$\cos ^{-1}\left(\frac{y}{b}\right)=n \log \left(\frac{x}{n}\right)$.Differentiating both sides with respect to $x$:$-\frac{1}{\sqrt{1-(y/b)^2}} \cdot \frac{1}{b} \cdot y_1 = n \cdot \frac{1}{(x/n)} \cdot \frac{1}{n}$Simplifying the expression:$-\frac{1}{\sqrt{(b^2-y^2)/b^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$$-\frac{b}{\sqrt{b^2-y^2}} \cdot \frac{y_1}{b} = \frac{n}{x}$$-\frac{y_1}{\sqrt{b^2-y^2}} = \frac{n}{x}$Cross-multiplying:$-x y_1 = n \sqrt{b^2-y^2}$$x y_1 + n \sqrt{b^2-y^2} = 0$.

If $\cos ^{-1}\left(\frac{y}{b}\right)=n \log \left(\frac{x}{n}\right)$,then

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