A value of x satisfying the equation sin left | vedclass

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Question

(A) Let $\lambda = \cot^{-1}(1+x)$,then $\cot \lambda = 1+x$. From the right-angled triangle with base $(1+x)$ and perpendicular $1$,the hypotenuse is

Vedclass · Accepted Answer

$-\frac{1}{2}$

Vedclass · Answer

(A) Let $\lambda = \cot^{-1}(1+x)$,then $\cot \lambda = 1+x$. From the right-angled triangle with base $(1+x)$ and perpendicular $1$,the hypotenuse is $\sqrt{(1+x)^2 + 1^2} = \sqrt{x^2 + 2x + 2}$. Thus,$\sin \lambda = \frac{1}{\sqrt{x^2 + 2x + 2}}$.Let $\beta = 	an^{-1}x$,then $	an \beta = x$. From the right-angled triangle with perpendicular $x$ and base $1$,the hypotenuse is $\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$. Thus,$\cos \beta = \frac{1}{\sqrt{x^2 + 1}}$.Given the equation $\sin \lambda = \cos \beta$,we have:$\frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{x^2 + 1}}$Squaring both sides:$x^2 + 2x + 2 = x^2 + 1$Subtracting $x^2$ from both sides:$2x + 2 = 1$$2x = -1$$x = -\frac{1}{2}$

$A$ value of $x$ satisfying the equation $\sin \left[ \cot^{-1} (1 + x) \right] = \cos \left[ \tan^{-1} x \right]$ is

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