Question

For all real numbers \(x\), except \(x = 0\) and \(x = 1\), the function \(F\) is defined by \[ F\left(\frac{x}{x-1}\right) = \frac{1}{x}. \] If \(0 < \alpha < 90^{\circ}\), then \(F((\text{cosec}\,\alpha)^2) =\)

• A. \((\sin \alpha)^2\)
• B. \((\cos \alpha)^2\)
• C. \((\tan \alpha)^2\)
• D. \((\cot \alpha)^2\)

Hint:

Solve \(y = x/(x-1)\) for \(x\) to get a general rule \(F(y) = 1 - 1/y\), then substitute \(y = \text{cosec}^2\alpha\) and simplify with \(1-\sin^2\alpha=\cos^2\alpha\).

Answer 1

The Correct Option is B

Step 1: Match the input directly.
We need $F((\text{cosec}\,\alpha)^2)$. Instead of finding a general formula for $F$, look for the value of $x$ that makes $\dfrac{x}{x-1}$ equal to $(\text{cosec}\,\alpha)^2$ itself, since we already know $F\left(\dfrac{x}{x-1}\right) = \dfrac{1}{x}$ for that same $x$.

Step 2: Set up the equation.
\[ \frac{x}{x-1} = \text{cosec}^2\alpha \]
Cross multiply: $x = \text{cosec}^2\alpha \,(x - 1)$, so $x = x\,\text{cosec}^2\alpha - \text{cosec}^2\alpha$.

Step 3: Solve for x.
\[ \text{cosec}^2\alpha = x(\text{cosec}^2\alpha - 1) \implies x = \frac{\text{cosec}^2\alpha}{\text{cosec}^2\alpha - 1} \]
Using the identity $\text{cosec}^2\alpha - 1 = \cot^2\alpha$, this becomes
\[ x = \frac{\text{cosec}^2\alpha}{\cot^2\alpha} \]

Step 4: Simplify x using sine and cosine.
$\text{cosec}^2\alpha = \dfrac{1}{\sin^2\alpha}$ and $\cot^2\alpha = \dfrac{\cos^2\alpha}{\sin^2\alpha}$, so
\[ x = \frac{1/\sin^2\alpha}{\cos^2\alpha/\sin^2\alpha} = \frac{1}{\cos^2\alpha} = \sec^2\alpha \]

Step 5: Get F as 1/x.
Since $F\left(\dfrac{x}{x-1}\right) = \dfrac{1}{x}$ and we built $x$ so that $\dfrac{x}{x-1} = \text{cosec}^2\alpha$,
\[ F(\text{cosec}^2\alpha) = \frac{1}{x} = \frac{1}{\sec^2\alpha} = \cos^2\alpha \]

Step 6: Final answer.
\[ \boxed{(\cos\alpha)^2} \]