If x=sqrt{1-tan y}, then frac{dy}{dx}= | vedclass

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(B) Given $x = \sqrt{1-	an y}$.Squaring both sides, we get $x^2 = 1 - 	an y$.Rearranging the terms, $	an y = 1 - x^2$.Now, differentiate both sides

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$-\frac{2x}{x^4-2x^2+2}$

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(B) Given $x = \sqrt{1-	an y}$.Squaring both sides, we get $x^2 = 1 - 	an y$.Rearranging the terms, $	an y = 1 - x^2$.Now, differentiate both sides with respect to $x$:$\frac{d}{dx}(	an y) = \frac{d}{dx}(1 - x^2)$.Using the chain rule, $\sec^2 y \cdot \frac{dy}{dx} = -2x$.We know that $\sec^2 y = 1 + 	an^2 y$.Substitute $	an y = 1 - x^2$ into the expression:$\sec^2 y = 1 + (1 - x^2)^2 = 1 + (1 - 2x^2 + x^4) = x^4 - 2x^2 + 2$.Now, substitute this back into the derivative equation:$(x^4 - 2x^2 + 2) \cdot \frac{dy}{dx} = -2x$.Therefore, $\frac{dy}{dx} = -\frac{2x}{x^4 - 2x^2 + 2}$.

If $x=\sqrt{1-\tan y}$, then $\frac{dy}{dx}=$

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