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: Rewrite F in terms of its own input.
We are told \(F\left(\dfrac{x}{x-1}\right) = \dfrac{1}{x}\). To find a general rule for \(F(y)\), let \(y = \dfrac{x}{x-1}\) and solve this for \(x\) in terms of \(y\).

Step 2: Solve for x.
\[ y(x - 1) = x \implies yx - y = x \implies yx - x = y \implies x(y - 1) = y \implies x = \frac{y}{y-1} \]

Step 3: Substitute back to get F(y).
Since \(F(y) = \dfrac{1}{x}\),
\[ F(y) = \frac{y-1}{y} = 1 - \frac{1}{y} \]
This is now a formula for \(F\) that works for any valid input \(y\), not just \(x/(x-1)\).

Step 4: Plug in \(y = (\text{cosec}\,\alpha)^2\).
\[ F((\text{cosec}\,\alpha)^2) = 1 - \frac{1}{(\text{cosec}\,\alpha)^2} = 1 - \sin^2 \alpha \]
because \(\dfrac{1}{\text{cosec}\,\alpha} = \sin \alpha\).

Step 5: Simplify using the Pythagorean identity.
\[ 1 - \sin^2 \alpha = \cos^2 \alpha \]

Step 6: Final Answer.
\[ \boxed{F((\text{cosec}\,\alpha)^2) = (\cos \alpha)^2} \]