If $$x = \sqrt{2^{\text{cosec}^{-1}t}}$$ and $$y = \sqrt{2^{\text{sec}^{-1}t}}$$, ($$|t| \geq 1$$), then $$\frac{dy}{dx}$$ is equal to:

We have two functions of the same independent variable $$t$$, namely

$$x=\sqrt{2^{\cosec^{-1}t}}\quad\text{and}\quad y=\sqrt{2^{\sec^{-1}t}},\qquad |t|\ge 1.$$

Our aim is to determine $$\dfrac{dy}{dx}$$. Because both $$x$$ and $$y$$ are written in terms of $$t$$, we shall first find $$\dfrac{dx}{dt}$$ and $$\dfrac{dy}{dt}$$ separately and then divide them, using the chain-rule identity

$$\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}.$$

Let us begin with $$x$$. Rewrite the square root as a rational exponent:

$$x=\Bigl(2^{\cosec^{-1}t}\Bigr)^{1/2}=2^{\tfrac12\,\cosec^{-1}t}.$$

Taking natural logarithms on both sides gives

$$\ln x=\frac12\,\cosec^{-1}t\;\ln 2.$$

Differentiating this equation with respect to $$t$$ we obtain

$$\frac1x\,\frac{dx}{dt}=\frac{\ln 2}{2}\,\frac{d}{dt}\bigl(\cosec^{-1}t\bigr).$$

We now recall the standard derivative formula for the inverse cosecant:

$$\frac{d}{dt}\bigl(\cosec^{-1}t\bigr)=-\frac1{|t|\sqrt{t^{2}-1}}.$$

Substituting this value, we get

$$\frac1x\,\frac{dx}{dt}=\frac{\ln 2}{2}\left(-\frac1{|t|\sqrt{t^{2}-1}}\right).$$

Multiplying both sides by $$x$$ yields

$$\frac{dx}{dt}=x\;\frac{\ln 2}{2}\left(-\frac1{|t|\sqrt{t^{2}-1}}\right).$$

We carry out exactly the same procedure for $$y$$. First rewrite:

$$y=\Bigl(2^{\sec^{-1}t}\Bigr)^{1/2}=2^{\tfrac12\,\sec^{-1}t},$$

and hence

$$\ln y=\frac12\,\sec^{-1}t\;\ln 2.$$

Differentiating with respect to $$t$$ gives

$$\frac1y\,\frac{dy}{dt}=\frac{\ln 2}{2}\,\frac{d}{dt}\bigl(\sec^{-1}t\bigr).$$

The derivative formula for the inverse secant is

$$\frac{d}{dt}\bigl(\sec^{-1}t\bigr)=\frac1{|t|\sqrt{t^{2}-1}}.$$

Substituting this, we find

$$\frac1y\,\frac{dy}{dt}=\frac{\ln 2}{2}\left(\frac1{|t|\sqrt{t^{2}-1}}\right).$$

Therefore

$$\frac{dy}{dt}=y\;\frac{\ln 2}{2}\left(\frac1{|t|\sqrt{t^{2}-1}}\right).$$

Now we form the desired ratio:

$$\frac{dy}{dx}=\frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}=\frac{y\,\dfrac{\ln 2}{2}\left(\dfrac1{|t|\sqrt{t^{2}-1}}\right)}{x\,\dfrac{\ln 2}{2}\left(-\dfrac1{|t|\sqrt{t^{2}-1}}\right)}.$$

Notice that the factors $$\dfrac{\ln 2}{2}$$, $$|t|$$ and $$\sqrt{t^{2}-1}$$ appear in both numerator and denominator; they cancel out completely, leaving only the minus sign and the ratio of $$y$$ to $$x$$:

$$\frac{dy}{dx}=-\frac{y}{x}.$$

Hence, the correct answer is Option B.