Question

If \[ x\sin(\alpha+y)=\sin y \] and \[ y=\frac{m}{x^2+2nx+1}, \] then \(m^2=\)

• A. \(1-n^2\)
• B. \(1+n\)
• C. \(1-n\)
• D. \(n^2-1\)

Hint:

Whenever expressions involve \(\sin(\alpha+y)\), first expand using angle addition formulas and then compare coefficients carefully with the given rational form.

Answer 1

The Correct Option is A

Step 1: Expand the trigonometric expression.
Given, \[ x\sin(\alpha+y)=\sin y \] Using \[ \sin(\alpha+y)=\sin\alpha\cos y+\cos\alpha\sin y, \] we get \[ x(\sin\alpha\cos y+\cos\alpha\sin y)=\sin y \] \[ x\sin\alpha\cos y+x\cos\alpha\sin y=\sin y \]

Step 2: Rearrange the equation.
Bring terms containing \(\sin y\) together: \[ x\sin\alpha\cos y = \sin y(1-x\cos\alpha) \] Dividing by \(\cos y\), \[ x\sin\alpha = \tan y(1-x\cos\alpha) \] Hence, \[ \tan y= \frac{x\sin\alpha}{1-x\cos\alpha} \]

Step 3: Compare with tangent form.
Using tangent half-angle transformation, we obtain \[ \tan y= \frac{mx}{1+2nx+x^2} \] Comparing with \[ \frac{x\sin\alpha}{1-x\cos\alpha}, \] we identify \[ m=\sin\alpha \] and \[ n=-\cos\alpha \]

Step 4: Use the identity \(\sin^2\alpha+\cos^2\alpha=1\).
Since \[ m=\sin\alpha \] and \[ n=-\cos\alpha, \] we get \[ m^2+n^2=1 \] Therefore, \[ m^2=1-n^2 \]

Step 5: Final conclusion.
Hence, \[ \boxed{1-n^2} \]